Mixing
Some basic questions regarding rank-1 matrices
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If an $ntimes n$ matrix $B$ has rank 1, and A is another $ntimes n$ matrix, then why does $AB$ also have rank 1? This showed up in a solution that I read through, but it doesn't seem like an obvious fact.
And one more thing that came up in this solution: it says that since this matrix has rank 1, then it must have $(n-1)$ eigenvalues that are all zero, and only one non-zero eigenvalue. I don't see how this has to be true either.
Any ideas are welcome.
Thanks,
linear-algebra matrices eigenvalues-eigenvectors matrix-rank
edited Dec 12 '15 at 20:43
Rory Daulton
29.4k53254
asked Dec 12 '15 at 14:22
User001User001
1
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add a comment |
$begingroup$
If an $ntimes n$ matrix $B$ has rank 1, and A is another $ntimes n$ matrix, then why does $AB$ also have rank 1? This showed up in a solution that I read through, but it doesn't seem like an obvious fact.
And one more thing that came up in this solution: it says that since this matrix has rank 1, then it must have $(n-1)$ eigenvalues that are all zero, and only one non-zero eigenvalue. I don't see how this has to be true either.
Any ideas are welcome.
Thanks,
linear-algebra matrices eigenvalues-eigenvectors matrix-rank
edited Dec 12 '15 at 20:43
Rory Daulton
29.4k53254
asked Dec 12 '15 at 14:22
User001User001
1
$endgroup$
add a comment |
$begingroup$
If an $ntimes n$ matrix $B$ has rank 1, and A is another $ntimes n$ matrix, then why does $AB$ also have rank 1? This showed up in a solution that I read through, but it doesn't seem like an obvious fact.
And one more thing that came up in this solution: it says that since this matrix has rank 1, then it must have $(n-1)$ eigenvalues that are all zero, and only one non-zero eigenvalue. I don't see how this has to be true either.
Any ideas are welcome.
Thanks,
linear-algebra matrices eigenvalues-eigenvectors matrix-rank
edited Dec 12 '15 at 20:43
Rory Daulton
29.4k53254
asked Dec 12 '15 at 14:22
User001User001
1
$endgroup$
If an $ntimes n$ matrix $B$ has rank 1, and A is another $ntimes n$ matrix, then why does $AB$ also have rank 1? This showed up in a solution that I read through, but it doesn't seem like an obvious fact.
And one more thing that came up in this solution: it says that since this matrix has rank 1, then it must have $(n-1)$ eigenvalues that are all zero, and only one non-zero eigenvalue. I don't see how this has to be true either.
Any ideas are welcome.
Thanks,
linear-algebra matrices eigenvalues-eigenvectors matrix-rank
linear-algebra matrices eigenvalues-eigenvectors matrix-rank
edited Dec 12 '15 at 20:43
Rory Daulton
29.4k53254
asked Dec 12 '15 at 14:22
User001User001
1
edited Dec 12 '15 at 20:43
Rory Daulton
29.4k53254
asked Dec 12 '15 at 14:22
User001User001
1
edited Dec 12 '15 at 20:43
Rory Daulton
29.4k53254
edited Dec 12 '15 at 20:43
Rory Daulton
29.4k53254
edited Dec 12 '15 at 20:43
Rory Daulton
29.4k53254
29.4k53254
asked Dec 12 '15 at 14:22
User001User001
1
asked Dec 12 '15 at 14:22
User001User001
1
asked Dec 12 '15 at 14:22
User001User001
1
1
add a comment |
add a comment |
4 Answers
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Okay, we can show this in different ways, depending on how we define rank. I'll try to show these using both these definitions:
$operatorname{rank}(A)$ is the dimension of the range of $A$.
$operatorname{rank}(A)$ is the least number $r$ of rank 1 matrices $A_1, A_2, dots, A_r$ so that $$A = A_1 + A_2 + dots + A_r$$ and a rank one matrix $A_1$ is defined as a matrix that can be written as an outer product: $A_1 = xy^T$ for vectors $x, y$.
These two definitions are equivalent. The proof of this is left as an exercise to the reader.
If $B$ has rank 1, then $AB$ has at most rank one
Fixed the formulation for you on this one. $AB$ can have rank zero.
Definition 1
$B$ has rank one, so its range is one dimensional. It follows that its nullspace is $n-1$-dimensional. Consider the range of $AB$. If $x$ is in the nullspace of $B$, then $ABx = A0 = 0$, so $AB$'s nullspace is also at least $n-1$-dimensional, so its range is at most 1-dimensional.
Note that for $AB$ to have rank 1 you must have that $Ax neq 0$ for some $x$ in the range of $B$. A sufficient, but not necessary, condition for this is that $A$ is invertible.
Definition 2
$B$ has rank one, so it can be written $B = xy^T$. Then $AB = Axy^T = (Ax)y^T$, so $AB$ can be written as one outer product (but we don't know if it can be written using zero outer products, which would be the case if $Ax = 0$).
A rank one matrix has $n-1$ zero eigenvalues
Definition 1
As said before, if $B$ has rank 1, then its nullspace is $n-1$ dimensional. Pick a basis $v_1, dots, v_{n-1}$ for the nullspace of $B$. These are eigenvectors of $B$ with eigenvalue zero.
Definition 2
Say $B = xy^T$. Then if you multiply a vector $v$ with $B$ you get $xy^Tv$. Note that $y^Tv$ is the inner product of $y$ and $v$. The orthogonal complement $mathcal U$ to the subspace spanned by $y$ has dimension $n-1$. If $u in mathcal U$ then $y^Tu = 0$ and hence
$$Bu = xy^Tu = x cdot 0 = 0$$
so if you pick a basis for $mathcal U$, this basis will be eigenvectors to $B$ with eigenvalue 0.
Is there a non-zero eigenvalue?
In short, it is not guaranteed that there will be a non-zero eigenvalue.
This rank one nilpotent matrix:
$$begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$$
has zero as eigenvalue with algebraic multiplicity 2 and geometric multiplicity 1. It therefore has $n-1$ zero eigenvalues, but it does not have 1 non-zero eigenvalue. In other words, it is not guaranteed that you will have a non-zero eigenvalue for rank one matrices.
answered Dec 12 '15 at 14:41
CalleCalle
6,37412442
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1
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@User001, yes, it would, I mentioned this under "Definition 1" under the section dealing with $AB$ having rank at most one.
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– Calle
Dec 12 '15 at 15:18
1
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@User001, since $B$ has rank one, there exists an $x$ such that $Bx = y neq 0$. If $A$ is invertible we know that $Ay neq 0$, so $AB$ has at least rank one, but we know that $AB$ has at most rank one, so $1 leq operatorname{rank}(AB) leq 1$ so the rank is one. If $A$ would not be invertible, we would not know that $Ay neq 0$.
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– Calle
Dec 12 '15 at 15:40
1
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Sorry, I meant $AB$. :)
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– Calle
Dec 12 '15 at 15:43
1
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Ok, got it. Thanks so much for your truly awesome solution, additional comments, and especially making many of us on this page aware of our mistake of assuming that a rank-1 matrix had to have a non-zero eigenvalue. I really appreciate it. Have a great night @Calle :-)
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– User001
Dec 12 '15 at 15:53
1
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@User001, Thank you for asking a question I enjoyed answering. :)
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– Calle
Dec 12 '15 at 15:54
|
show 8 more comments
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It comes from the associavity of matrix multiplication. If $B$ has rank-1 then it can be written in the form $B = u v^T$ for some vectors $u$ and $v$. So,
$$AB = A (uv^T) = (Au) v^T$$
but $Au$ is a vector itself, so now we have a rank-1 expression for $AB$. Maybe it will be helpful to see a picture of the shapes of the matrices during this process:
$$underbrace{begin{bmatrix}cdot & cdot & cdot \ cdot & cdot & cdot \ cdot & cdot & cdotend{bmatrix}}_{A} left(underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{u}underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T}right)=
left(underbrace{begin{bmatrix}cdot & cdot & cdot \ cdot & cdot & cdot \ cdot & cdot & cdotend{bmatrix}}_{A}underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{u}right)underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T} =
underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{Au}underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T} $$
(Of course, note that if $Au=0$, then the result is the zero matrix so the rank would be zero instead of 1)
answered Dec 12 '15 at 22:23
Nick AlgerNick Alger
9,85863267
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What if $Au = 0$?
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– Jacob Maibach
Dec 12 '15 at 22:31
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Then the rank is zero and the statement in the original question is false, obviously...
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– Nick Alger
Dec 12 '15 at 22:33
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I was pointing out that your approach seems to indicate that the rank of $AB$ is always 1, or at least that is how I read it.
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– Jacob Maibach
Dec 12 '15 at 22:47
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@JacobMaibach Ok, I added a sentence to explicitly mention this.
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– Nick Alger
Dec 13 '15 at 1:19
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Hi @NickAlger, your answer is awesome. So simple and intuitive to understand. Thanks so much :-)
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– User001
Dec 13 '15 at 5:29
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Concerning the first point, remember $A$ and $B$ represent endomorphisms $f$ and $g$ in some basis, and $AB$ represent $fcirc g$.
Now $;DeclareMathOperatorrk{rank}DeclareMathOperatorimg{Im}rk B=1$ means $;dimimg g=1$. Then we know $;dimimg(fcirc g)ledimimg g$, which means $;rk ABlerk B$. So $rk AB=0$ or $1$.
For the second point, if $AB$ has indeed rank $1$, its kernel has dimension $n-1$. As the kernel is the eigenspace for the eigenvalue $0$, and has dimension at most equal to the multiplicity of this eigenvalues, it implies the eigenvalue $0$ has multiplicity at least $n-1$, and it can't have multiplicity $n$ as it would imply $rk AB=0$.
answered Dec 12 '15 at 14:38
BernardBernard
119k740113
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Hi @Bernard, I apologize for probably withholding a key piece of information that I just found from rereading that solution. A is invertible. Does that now make $AB$ have rank 1, for certain? Thanks,
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– User001
Dec 12 '15 at 14:41
2
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Yes: if $A$ is invertible, $operatorname{rank}AB=operatorname{rank}B$, and similarly , $operatorname{rank}BA=operatorname{rank}B$.
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– Bernard
Dec 12 '15 at 19:16
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The rank of a matrix is the dimension of its image ${rm rank} B = dim {Bx: x in mathbb R^n},$ or equivalently the dimension of the space spanned by its columns.
Since applying $AB$ on $mathbb R^n$ is the same as applying $A$ on the one-dimensional space $Bmathbb R^n$, we see that $AB$ can only have at most rank 1. Note that rank 0 is possible when the columns of $B$ are orthogonal to the rows of $A$.
We have the rank-nullity theorem $n = {rm rank} B + dim ker B$ (i.e. what doesn't contribute to the dimensions of the image must land in the kernel.
If ${rm rank} B = 1$, then the dimension of the kernel must be $n-1$, i.e. the eigenspace for the eigenvalue is $n-1$. We can deduce that there is also another nonzero eigenvalue since otherwise we'd have ${rm rank} AB = 0$.
answered Dec 12 '15 at 14:35
RolandRoland
4,80211331
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Hi @Roland, the matrix $A$ is actually assumed to be invertible. I reread the solution and noticed that. Does the invertibility of $A$ make $AB$ have rank 1, for certain? Thanks so much,
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– User001
Dec 12 '15 at 14:58
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"We can deduce that there is also another nonzero eigenvalue since otherwise we'd have rank B = 0." This is not true.
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– Calle
Dec 12 '15 at 15:14
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@User001: Yes, because then $A^{-1}(AB) = B$ has rank 1, and so $AB$ must have rank at least 1.
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– Ilmari Karonen
Dec 12 '15 at 15:23
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HI @IlmariKaronen, how does your equation imply that $AB$ must have rank at least 1? It seems like you are using the invertibility of $A^{-1}$, but I'm not sure how...thanks,
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– User001
Dec 12 '15 at 15:28
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@User001: By the same argument given by Roland above, the product of two matrices cannot have a higher rank than either of the matrices alone. Since we know that $A^{-1}(AB) = (A^{-1}A)B = B$ has rank 1, this means that both $A^{-1}$ and $AB$ must have rank at least 1. (Of course, for $A^{-1}$, we actually know more; namely that, as invertible matrices, both $A$ and $A^{-1}$ in fact have full rank. In fact, we can prove this simply by replacing $B$ in the previous argument with, say, the identity matrix. But for the proof that $AB$ has rank 1, we don't really need this extra information.)
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– Ilmari Karonen
Dec 12 '15 at 16:04
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4 Answers
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$begingroup$
Okay, we can show this in different ways, depending on how we define rank. I'll try to show these using both these definitions:
$operatorname{rank}(A)$ is the dimension of the range of $A$.
$operatorname{rank}(A)$ is the least number $r$ of rank 1 matrices $A_1, A_2, dots, A_r$ so that $$A = A_1 + A_2 + dots + A_r$$ and a rank one matrix $A_1$ is defined as a matrix that can be written as an outer product: $A_1 = xy^T$ for vectors $x, y$.
These two definitions are equivalent. The proof of this is left as an exercise to the reader.
If $B$ has rank 1, then $AB$ has at most rank one
Fixed the formulation for you on this one. $AB$ can have rank zero.
Definition 1
$B$ has rank one, so its range is one dimensional. It follows that its nullspace is $n-1$-dimensional. Consider the range of $AB$. If $x$ is in the nullspace of $B$, then $ABx = A0 = 0$, so $AB$'s nullspace is also at least $n-1$-dimensional, so its range is at most 1-dimensional.
Note that for $AB$ to have rank 1 you must have that $Ax neq 0$ for some $x$ in the range of $B$. A sufficient, but not necessary, condition for this is that $A$ is invertible.
Definition 2
$B$ has rank one, so it can be written $B = xy^T$. Then $AB = Axy^T = (Ax)y^T$, so $AB$ can be written as one outer product (but we don't know if it can be written using zero outer products, which would be the case if $Ax = 0$).
A rank one matrix has $n-1$ zero eigenvalues
Definition 1
As said before, if $B$ has rank 1, then its nullspace is $n-1$ dimensional. Pick a basis $v_1, dots, v_{n-1}$ for the nullspace of $B$. These are eigenvectors of $B$ with eigenvalue zero.
Definition 2
Say $B = xy^T$. Then if you multiply a vector $v$ with $B$ you get $xy^Tv$. Note that $y^Tv$ is the inner product of $y$ and $v$. The orthogonal complement $mathcal U$ to the subspace spanned by $y$ has dimension $n-1$. If $u in mathcal U$ then $y^Tu = 0$ and hence
$$Bu = xy^Tu = x cdot 0 = 0$$
so if you pick a basis for $mathcal U$, this basis will be eigenvectors to $B$ with eigenvalue 0.
Is there a non-zero eigenvalue?
In short, it is not guaranteed that there will be a non-zero eigenvalue.
This rank one nilpotent matrix:
$$begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$$
has zero as eigenvalue with algebraic multiplicity 2 and geometric multiplicity 1. It therefore has $n-1$ zero eigenvalues, but it does not have 1 non-zero eigenvalue. In other words, it is not guaranteed that you will have a non-zero eigenvalue for rank one matrices.
answered Dec 12 '15 at 14:41
CalleCalle
6,37412442
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1
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@User001, yes, it would, I mentioned this under "Definition 1" under the section dealing with $AB$ having rank at most one.
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– Calle
Dec 12 '15 at 15:18
1
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@User001, since $B$ has rank one, there exists an $x$ such that $Bx = y neq 0$. If $A$ is invertible we know that $Ay neq 0$, so $AB$ has at least rank one, but we know that $AB$ has at most rank one, so $1 leq operatorname{rank}(AB) leq 1$ so the rank is one. If $A$ would not be invertible, we would not know that $Ay neq 0$.
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– Calle
Dec 12 '15 at 15:40
1
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Sorry, I meant $AB$. :)
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– Calle
Dec 12 '15 at 15:43
1
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Ok, got it. Thanks so much for your truly awesome solution, additional comments, and especially making many of us on this page aware of our mistake of assuming that a rank-1 matrix had to have a non-zero eigenvalue. I really appreciate it. Have a great night @Calle :-)
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– User001
Dec 12 '15 at 15:53
1
$begingroup$
@User001, Thank you for asking a question I enjoyed answering. :)
$endgroup$
– Calle
Dec 12 '15 at 15:54
|
show 8 more comments
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Okay, we can show this in different ways, depending on how we define rank. I'll try to show these using both these definitions:
$operatorname{rank}(A)$ is the dimension of the range of $A$.
$operatorname{rank}(A)$ is the least number $r$ of rank 1 matrices $A_1, A_2, dots, A_r$ so that $$A = A_1 + A_2 + dots + A_r$$ and a rank one matrix $A_1$ is defined as a matrix that can be written as an outer product: $A_1 = xy^T$ for vectors $x, y$.
These two definitions are equivalent. The proof of this is left as an exercise to the reader.
If $B$ has rank 1, then $AB$ has at most rank one
Fixed the formulation for you on this one. $AB$ can have rank zero.
Definition 1
$B$ has rank one, so its range is one dimensional. It follows that its nullspace is $n-1$-dimensional. Consider the range of $AB$. If $x$ is in the nullspace of $B$, then $ABx = A0 = 0$, so $AB$'s nullspace is also at least $n-1$-dimensional, so its range is at most 1-dimensional.
Note that for $AB$ to have rank 1 you must have that $Ax neq 0$ for some $x$ in the range of $B$. A sufficient, but not necessary, condition for this is that $A$ is invertible.
Definition 2
$B$ has rank one, so it can be written $B = xy^T$. Then $AB = Axy^T = (Ax)y^T$, so $AB$ can be written as one outer product (but we don't know if it can be written using zero outer products, which would be the case if $Ax = 0$).
A rank one matrix has $n-1$ zero eigenvalues
Definition 1
As said before, if $B$ has rank 1, then its nullspace is $n-1$ dimensional. Pick a basis $v_1, dots, v_{n-1}$ for the nullspace of $B$. These are eigenvectors of $B$ with eigenvalue zero.
Definition 2
Say $B = xy^T$. Then if you multiply a vector $v$ with $B$ you get $xy^Tv$. Note that $y^Tv$ is the inner product of $y$ and $v$. The orthogonal complement $mathcal U$ to the subspace spanned by $y$ has dimension $n-1$. If $u in mathcal U$ then $y^Tu = 0$ and hence
$$Bu = xy^Tu = x cdot 0 = 0$$
so if you pick a basis for $mathcal U$, this basis will be eigenvectors to $B$ with eigenvalue 0.
Is there a non-zero eigenvalue?
In short, it is not guaranteed that there will be a non-zero eigenvalue.
This rank one nilpotent matrix:
$$begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$$
has zero as eigenvalue with algebraic multiplicity 2 and geometric multiplicity 1. It therefore has $n-1$ zero eigenvalues, but it does not have 1 non-zero eigenvalue. In other words, it is not guaranteed that you will have a non-zero eigenvalue for rank one matrices.
answered Dec 12 '15 at 14:41
CalleCalle
6,37412442
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1
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@User001, yes, it would, I mentioned this under "Definition 1" under the section dealing with $AB$ having rank at most one.
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– Calle
Dec 12 '15 at 15:18
1
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@User001, since $B$ has rank one, there exists an $x$ such that $Bx = y neq 0$. If $A$ is invertible we know that $Ay neq 0$, so $AB$ has at least rank one, but we know that $AB$ has at most rank one, so $1 leq operatorname{rank}(AB) leq 1$ so the rank is one. If $A$ would not be invertible, we would not know that $Ay neq 0$.
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– Calle
Dec 12 '15 at 15:40
1
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Sorry, I meant $AB$. :)
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– Calle
Dec 12 '15 at 15:43
1
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Ok, got it. Thanks so much for your truly awesome solution, additional comments, and especially making many of us on this page aware of our mistake of assuming that a rank-1 matrix had to have a non-zero eigenvalue. I really appreciate it. Have a great night @Calle :-)
$endgroup$
– User001
Dec 12 '15 at 15:53
1
$begingroup$
@User001, Thank you for asking a question I enjoyed answering. :)
$endgroup$
– Calle
Dec 12 '15 at 15:54
|
show 8 more comments
$begingroup$
Okay, we can show this in different ways, depending on how we define rank. I'll try to show these using both these definitions:
$operatorname{rank}(A)$ is the dimension of the range of $A$.
$operatorname{rank}(A)$ is the least number $r$ of rank 1 matrices $A_1, A_2, dots, A_r$ so that $$A = A_1 + A_2 + dots + A_r$$ and a rank one matrix $A_1$ is defined as a matrix that can be written as an outer product: $A_1 = xy^T$ for vectors $x, y$.
These two definitions are equivalent. The proof of this is left as an exercise to the reader.
If $B$ has rank 1, then $AB$ has at most rank one
Fixed the formulation for you on this one. $AB$ can have rank zero.
Definition 1
$B$ has rank one, so its range is one dimensional. It follows that its nullspace is $n-1$-dimensional. Consider the range of $AB$. If $x$ is in the nullspace of $B$, then $ABx = A0 = 0$, so $AB$'s nullspace is also at least $n-1$-dimensional, so its range is at most 1-dimensional.
Note that for $AB$ to have rank 1 you must have that $Ax neq 0$ for some $x$ in the range of $B$. A sufficient, but not necessary, condition for this is that $A$ is invertible.
Definition 2
$B$ has rank one, so it can be written $B = xy^T$. Then $AB = Axy^T = (Ax)y^T$, so $AB$ can be written as one outer product (but we don't know if it can be written using zero outer products, which would be the case if $Ax = 0$).
A rank one matrix has $n-1$ zero eigenvalues
Definition 1
As said before, if $B$ has rank 1, then its nullspace is $n-1$ dimensional. Pick a basis $v_1, dots, v_{n-1}$ for the nullspace of $B$. These are eigenvectors of $B$ with eigenvalue zero.
Definition 2
Say $B = xy^T$. Then if you multiply a vector $v$ with $B$ you get $xy^Tv$. Note that $y^Tv$ is the inner product of $y$ and $v$. The orthogonal complement $mathcal U$ to the subspace spanned by $y$ has dimension $n-1$. If $u in mathcal U$ then $y^Tu = 0$ and hence
$$Bu = xy^Tu = x cdot 0 = 0$$
so if you pick a basis for $mathcal U$, this basis will be eigenvectors to $B$ with eigenvalue 0.
Is there a non-zero eigenvalue?
In short, it is not guaranteed that there will be a non-zero eigenvalue.
This rank one nilpotent matrix:
$$begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$$
has zero as eigenvalue with algebraic multiplicity 2 and geometric multiplicity 1. It therefore has $n-1$ zero eigenvalues, but it does not have 1 non-zero eigenvalue. In other words, it is not guaranteed that you will have a non-zero eigenvalue for rank one matrices.
answered Dec 12 '15 at 14:41
CalleCalle
6,37412442
$endgroup$
Okay, we can show this in different ways, depending on how we define rank. I'll try to show these using both these definitions:
$operatorname{rank}(A)$ is the dimension of the range of $A$.
$operatorname{rank}(A)$ is the least number $r$ of rank 1 matrices $A_1, A_2, dots, A_r$ so that $$A = A_1 + A_2 + dots + A_r$$ and a rank one matrix $A_1$ is defined as a matrix that can be written as an outer product: $A_1 = xy^T$ for vectors $x, y$.
These two definitions are equivalent. The proof of this is left as an exercise to the reader.
If $B$ has rank 1, then $AB$ has at most rank one
Fixed the formulation for you on this one. $AB$ can have rank zero.
Definition 1
$B$ has rank one, so its range is one dimensional. It follows that its nullspace is $n-1$-dimensional. Consider the range of $AB$. If $x$ is in the nullspace of $B$, then $ABx = A0 = 0$, so $AB$'s nullspace is also at least $n-1$-dimensional, so its range is at most 1-dimensional.
Note that for $AB$ to have rank 1 you must have that $Ax neq 0$ for some $x$ in the range of $B$. A sufficient, but not necessary, condition for this is that $A$ is invertible.
Definition 2
$B$ has rank one, so it can be written $B = xy^T$. Then $AB = Axy^T = (Ax)y^T$, so $AB$ can be written as one outer product (but we don't know if it can be written using zero outer products, which would be the case if $Ax = 0$).
A rank one matrix has $n-1$ zero eigenvalues
Definition 1
As said before, if $B$ has rank 1, then its nullspace is $n-1$ dimensional. Pick a basis $v_1, dots, v_{n-1}$ for the nullspace of $B$. These are eigenvectors of $B$ with eigenvalue zero.
Definition 2
Say $B = xy^T$. Then if you multiply a vector $v$ with $B$ you get $xy^Tv$. Note that $y^Tv$ is the inner product of $y$ and $v$. The orthogonal complement $mathcal U$ to the subspace spanned by $y$ has dimension $n-1$. If $u in mathcal U$ then $y^Tu = 0$ and hence
$$Bu = xy^Tu = x cdot 0 = 0$$
so if you pick a basis for $mathcal U$, this basis will be eigenvectors to $B$ with eigenvalue 0.
Is there a non-zero eigenvalue?
In short, it is not guaranteed that there will be a non-zero eigenvalue.
This rank one nilpotent matrix:
$$begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$$
has zero as eigenvalue with algebraic multiplicity 2 and geometric multiplicity 1. It therefore has $n-1$ zero eigenvalues, but it does not have 1 non-zero eigenvalue. In other words, it is not guaranteed that you will have a non-zero eigenvalue for rank one matrices.
answered Dec 12 '15 at 14:41
CalleCalle
6,37412442
edited Jan 5 at 13:38
answered Dec 12 '15 at 14:41
CalleCalle
6,37412442
answered Dec 12 '15 at 14:41
CalleCalle
6,37412442
answered Dec 12 '15 at 14:41
CalleCalle
6,37412442
6,37412442
1
$begingroup$
@User001, yes, it would, I mentioned this under "Definition 1" under the section dealing with $AB$ having rank at most one.
$endgroup$
– Calle
Dec 12 '15 at 15:18
1
$begingroup$
@User001, since $B$ has rank one, there exists an $x$ such that $Bx = y neq 0$. If $A$ is invertible we know that $Ay neq 0$, so $AB$ has at least rank one, but we know that $AB$ has at most rank one, so $1 leq operatorname{rank}(AB) leq 1$ so the rank is one. If $A$ would not be invertible, we would not know that $Ay neq 0$.
$endgroup$
– Calle
Dec 12 '15 at 15:40
1
$begingroup$
Sorry, I meant $AB$. :)
$endgroup$
– Calle
Dec 12 '15 at 15:43
1
$begingroup$
Ok, got it. Thanks so much for your truly awesome solution, additional comments, and especially making many of us on this page aware of our mistake of assuming that a rank-1 matrix had to have a non-zero eigenvalue. I really appreciate it. Have a great night @Calle :-)
$endgroup$
– User001
Dec 12 '15 at 15:53
1
$begingroup$
@User001, Thank you for asking a question I enjoyed answering. :)
$endgroup$
– Calle
Dec 12 '15 at 15:54
|
show 8 more comments
1
$begingroup$
@User001, yes, it would, I mentioned this under "Definition 1" under the section dealing with $AB$ having rank at most one.
$endgroup$
– Calle
Dec 12 '15 at 15:18
1
$begingroup$
@User001, since $B$ has rank one, there exists an $x$ such that $Bx = y neq 0$. If $A$ is invertible we know that $Ay neq 0$, so $AB$ has at least rank one, but we know that $AB$ has at most rank one, so $1 leq operatorname{rank}(AB) leq 1$ so the rank is one. If $A$ would not be invertible, we would not know that $Ay neq 0$.
$endgroup$
– Calle
Dec 12 '15 at 15:40
1
$begingroup$
Sorry, I meant $AB$. :)
$endgroup$
– Calle
Dec 12 '15 at 15:43
1
$begingroup$
Ok, got it. Thanks so much for your truly awesome solution, additional comments, and especially making many of us on this page aware of our mistake of assuming that a rank-1 matrix had to have a non-zero eigenvalue. I really appreciate it. Have a great night @Calle :-)
$endgroup$
– User001
Dec 12 '15 at 15:53
1
$begingroup$
@User001, Thank you for asking a question I enjoyed answering. :)
$endgroup$
– Calle
Dec 12 '15 at 15:54
1
1
$begingroup$
@User001, yes, it would, I mentioned this under "Definition 1" under the section dealing with $AB$ having rank at most one.
$endgroup$
– Calle
Dec 12 '15 at 15:18
$begingroup$
@User001, yes, it would, I mentioned this under "Definition 1" under the section dealing with $AB$ having rank at most one.
$endgroup$
– Calle
Dec 12 '15 at 15:18
1
1
$begingroup$
@User001, since $B$ has rank one, there exists an $x$ such that $Bx = y neq 0$. If $A$ is invertible we know that $Ay neq 0$, so $AB$ has at least rank one, but we know that $AB$ has at most rank one, so $1 leq operatorname{rank}(AB) leq 1$ so the rank is one. If $A$ would not be invertible, we would not know that $Ay neq 0$.
$endgroup$
– Calle
Dec 12 '15 at 15:40
$begingroup$
@User001, since $B$ has rank one, there exists an $x$ such that $Bx = y neq 0$. If $A$ is invertible we know that $Ay neq 0$, so $AB$ has at least rank one, but we know that $AB$ has at most rank one, so $1 leq operatorname{rank}(AB) leq 1$ so the rank is one. If $A$ would not be invertible, we would not know that $Ay neq 0$.
$endgroup$
– Calle
Dec 12 '15 at 15:40
1
1
$begingroup$
Sorry, I meant $AB$. :)
$endgroup$
– Calle
Dec 12 '15 at 15:43
$begingroup$
Sorry, I meant $AB$. :)
$endgroup$
– Calle
Dec 12 '15 at 15:43
1
1
$begingroup$
Ok, got it. Thanks so much for your truly awesome solution, additional comments, and especially making many of us on this page aware of our mistake of assuming that a rank-1 matrix had to have a non-zero eigenvalue. I really appreciate it. Have a great night @Calle :-)
$endgroup$
– User001
Dec 12 '15 at 15:53
$begingroup$
Ok, got it. Thanks so much for your truly awesome solution, additional comments, and especially making many of us on this page aware of our mistake of assuming that a rank-1 matrix had to have a non-zero eigenvalue. I really appreciate it. Have a great night @Calle :-)
$endgroup$
– User001
Dec 12 '15 at 15:53
1
1
$begingroup$
@User001, Thank you for asking a question I enjoyed answering. :)
$endgroup$
– Calle
Dec 12 '15 at 15:54
$begingroup$
@User001, Thank you for asking a question I enjoyed answering. :)
$endgroup$
– Calle
Dec 12 '15 at 15:54
|
show 8 more comments
$begingroup$
It comes from the associavity of matrix multiplication. If $B$ has rank-1 then it can be written in the form $B = u v^T$ for some vectors $u$ and $v$. So,
$$AB = A (uv^T) = (Au) v^T$$
but $Au$ is a vector itself, so now we have a rank-1 expression for $AB$. Maybe it will be helpful to see a picture of the shapes of the matrices during this process:
$$underbrace{begin{bmatrix}cdot & cdot & cdot \ cdot & cdot & cdot \ cdot & cdot & cdotend{bmatrix}}_{A} left(underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{u}underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T}right)=
left(underbrace{begin{bmatrix}cdot & cdot & cdot \ cdot & cdot & cdot \ cdot & cdot & cdotend{bmatrix}}_{A}underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{u}right)underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T} =
underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{Au}underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T} $$
(Of course, note that if $Au=0$, then the result is the zero matrix so the rank would be zero instead of 1)
answered Dec 12 '15 at 22:23
Nick AlgerNick Alger
9,85863267
$endgroup$
$begingroup$
What if $Au = 0$?
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:31
$begingroup$
Then the rank is zero and the statement in the original question is false, obviously...
$endgroup$
– Nick Alger
Dec 12 '15 at 22:33
$begingroup$
I was pointing out that your approach seems to indicate that the rank of $AB$ is always 1, or at least that is how I read it.
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:47
$begingroup$
@JacobMaibach Ok, I added a sentence to explicitly mention this.
$endgroup$
– Nick Alger
Dec 13 '15 at 1:19
$begingroup$
Hi @NickAlger, your answer is awesome. So simple and intuitive to understand. Thanks so much :-)
$endgroup$
– User001
Dec 13 '15 at 5:29
add a comment |
$begingroup$
It comes from the associavity of matrix multiplication. If $B$ has rank-1 then it can be written in the form $B = u v^T$ for some vectors $u$ and $v$. So,
$$AB = A (uv^T) = (Au) v^T$$
but $Au$ is a vector itself, so now we have a rank-1 expression for $AB$. Maybe it will be helpful to see a picture of the shapes of the matrices during this process:
$$underbrace{begin{bmatrix}cdot & cdot & cdot \ cdot & cdot & cdot \ cdot & cdot & cdotend{bmatrix}}_{A} left(underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{u}underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T}right)=
left(underbrace{begin{bmatrix}cdot & cdot & cdot \ cdot & cdot & cdot \ cdot & cdot & cdotend{bmatrix}}_{A}underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{u}right)underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T} =
underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{Au}underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T} $$
(Of course, note that if $Au=0$, then the result is the zero matrix so the rank would be zero instead of 1)
answered Dec 12 '15 at 22:23
Nick AlgerNick Alger
9,85863267
$endgroup$
$begingroup$
What if $Au = 0$?
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:31
$begingroup$
Then the rank is zero and the statement in the original question is false, obviously...
$endgroup$
– Nick Alger
Dec 12 '15 at 22:33
$begingroup$
I was pointing out that your approach seems to indicate that the rank of $AB$ is always 1, or at least that is how I read it.
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:47
$begingroup$
@JacobMaibach Ok, I added a sentence to explicitly mention this.
$endgroup$
– Nick Alger
Dec 13 '15 at 1:19
$begingroup$
Hi @NickAlger, your answer is awesome. So simple and intuitive to understand. Thanks so much :-)
$endgroup$
– User001
Dec 13 '15 at 5:29
add a comment |
$begingroup$
It comes from the associavity of matrix multiplication. If $B$ has rank-1 then it can be written in the form $B = u v^T$ for some vectors $u$ and $v$. So,
$$AB = A (uv^T) = (Au) v^T$$
but $Au$ is a vector itself, so now we have a rank-1 expression for $AB$. Maybe it will be helpful to see a picture of the shapes of the matrices during this process:
$$underbrace{begin{bmatrix}cdot & cdot & cdot \ cdot & cdot & cdot \ cdot & cdot & cdotend{bmatrix}}_{A} left(underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{u}underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T}right)=
left(underbrace{begin{bmatrix}cdot & cdot & cdot \ cdot & cdot & cdot \ cdot & cdot & cdotend{bmatrix}}_{A}underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{u}right)underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T} =
underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{Au}underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T} $$
(Of course, note that if $Au=0$, then the result is the zero matrix so the rank would be zero instead of 1)
answered Dec 12 '15 at 22:23
Nick AlgerNick Alger
9,85863267
$endgroup$
It comes from the associavity of matrix multiplication. If $B$ has rank-1 then it can be written in the form $B = u v^T$ for some vectors $u$ and $v$. So,
$$AB = A (uv^T) = (Au) v^T$$
but $Au$ is a vector itself, so now we have a rank-1 expression for $AB$. Maybe it will be helpful to see a picture of the shapes of the matrices during this process:
$$underbrace{begin{bmatrix}cdot & cdot & cdot \ cdot & cdot & cdot \ cdot & cdot & cdotend{bmatrix}}_{A} left(underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{u}underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T}right)=
left(underbrace{begin{bmatrix}cdot & cdot & cdot \ cdot & cdot & cdot \ cdot & cdot & cdotend{bmatrix}}_{A}underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{u}right)underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T} =
underbrace{begin{bmatrix}cdot \ cdot \ cdotend{bmatrix}}_{Au}underbrace{begin{bmatrix}cdot & cdot & cdotend{bmatrix}}_{v^T} $$
(Of course, note that if $Au=0$, then the result is the zero matrix so the rank would be zero instead of 1)
answered Dec 12 '15 at 22:23
Nick AlgerNick Alger
9,85863267
edited Dec 13 '15 at 1:17
answered Dec 12 '15 at 22:23
Nick AlgerNick Alger
9,85863267
answered Dec 12 '15 at 22:23
Nick AlgerNick Alger
9,85863267
answered Dec 12 '15 at 22:23
Nick AlgerNick Alger
9,85863267
9,85863267
$begingroup$
What if $Au = 0$?
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:31
$begingroup$
Then the rank is zero and the statement in the original question is false, obviously...
$endgroup$
– Nick Alger
Dec 12 '15 at 22:33
$begingroup$
I was pointing out that your approach seems to indicate that the rank of $AB$ is always 1, or at least that is how I read it.
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:47
$begingroup$
@JacobMaibach Ok, I added a sentence to explicitly mention this.
$endgroup$
– Nick Alger
Dec 13 '15 at 1:19
$begingroup$
Hi @NickAlger, your answer is awesome. So simple and intuitive to understand. Thanks so much :-)
$endgroup$
– User001
Dec 13 '15 at 5:29
add a comment |
$begingroup$
What if $Au = 0$?
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:31
$begingroup$
Then the rank is zero and the statement in the original question is false, obviously...
$endgroup$
– Nick Alger
Dec 12 '15 at 22:33
$begingroup$
I was pointing out that your approach seems to indicate that the rank of $AB$ is always 1, or at least that is how I read it.
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:47
$begingroup$
@JacobMaibach Ok, I added a sentence to explicitly mention this.
$endgroup$
– Nick Alger
Dec 13 '15 at 1:19
$begingroup$
Hi @NickAlger, your answer is awesome. So simple and intuitive to understand. Thanks so much :-)
$endgroup$
– User001
Dec 13 '15 at 5:29
$begingroup$
What if $Au = 0$?
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:31
$begingroup$
What if $Au = 0$?
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:31
$begingroup$
Then the rank is zero and the statement in the original question is false, obviously...
$endgroup$
– Nick Alger
Dec 12 '15 at 22:33
$begingroup$
Then the rank is zero and the statement in the original question is false, obviously...
$endgroup$
– Nick Alger
Dec 12 '15 at 22:33
$begingroup$
I was pointing out that your approach seems to indicate that the rank of $AB$ is always 1, or at least that is how I read it.
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:47
$begingroup$
I was pointing out that your approach seems to indicate that the rank of $AB$ is always 1, or at least that is how I read it.
$endgroup$
– Jacob Maibach
Dec 12 '15 at 22:47
$begingroup$
@JacobMaibach Ok, I added a sentence to explicitly mention this.
$endgroup$
– Nick Alger
Dec 13 '15 at 1:19
$begingroup$
@JacobMaibach Ok, I added a sentence to explicitly mention this.
$endgroup$
– Nick Alger
Dec 13 '15 at 1:19
$begingroup$
Hi @NickAlger, your answer is awesome. So simple and intuitive to understand. Thanks so much :-)
$endgroup$
– User001
Dec 13 '15 at 5:29
$begingroup$
Hi @NickAlger, your answer is awesome. So simple and intuitive to understand. Thanks so much :-)
$endgroup$
– User001
Dec 13 '15 at 5:29
add a comment |
$begingroup$
Concerning the first point, remember $A$ and $B$ represent endomorphisms $f$ and $g$ in some basis, and $AB$ represent $fcirc g$.
Now $;DeclareMathOperatorrk{rank}DeclareMathOperatorimg{Im}rk B=1$ means $;dimimg g=1$. Then we know $;dimimg(fcirc g)ledimimg g$, which means $;rk ABlerk B$. So $rk AB=0$ or $1$.
For the second point, if $AB$ has indeed rank $1$, its kernel has dimension $n-1$. As the kernel is the eigenspace for the eigenvalue $0$, and has dimension at most equal to the multiplicity of this eigenvalues, it implies the eigenvalue $0$ has multiplicity at least $n-1$, and it can't have multiplicity $n$ as it would imply $rk AB=0$.
answered Dec 12 '15 at 14:38
BernardBernard
119k740113
$endgroup$
$begingroup$
Hi @Bernard, I apologize for probably withholding a key piece of information that I just found from rereading that solution. A is invertible. Does that now make $AB$ have rank 1, for certain? Thanks,
$endgroup$
– User001
Dec 12 '15 at 14:41
2
$begingroup$
Yes: if $A$ is invertible, $operatorname{rank}AB=operatorname{rank}B$, and similarly , $operatorname{rank}BA=operatorname{rank}B$.
$endgroup$
– Bernard
Dec 12 '15 at 19:16
add a comment |
$begingroup$
Concerning the first point, remember $A$ and $B$ represent endomorphisms $f$ and $g$ in some basis, and $AB$ represent $fcirc g$.
Now $;DeclareMathOperatorrk{rank}DeclareMathOperatorimg{Im}rk B=1$ means $;dimimg g=1$. Then we know $;dimimg(fcirc g)ledimimg g$, which means $;rk ABlerk B$. So $rk AB=0$ or $1$.
For the second point, if $AB$ has indeed rank $1$, its kernel has dimension $n-1$. As the kernel is the eigenspace for the eigenvalue $0$, and has dimension at most equal to the multiplicity of this eigenvalues, it implies the eigenvalue $0$ has multiplicity at least $n-1$, and it can't have multiplicity $n$ as it would imply $rk AB=0$.
answered Dec 12 '15 at 14:38
BernardBernard
119k740113
$endgroup$
$begingroup$
Hi @Bernard, I apologize for probably withholding a key piece of information that I just found from rereading that solution. A is invertible. Does that now make $AB$ have rank 1, for certain? Thanks,
$endgroup$
– User001
Dec 12 '15 at 14:41
2
$begingroup$
Yes: if $A$ is invertible, $operatorname{rank}AB=operatorname{rank}B$, and similarly , $operatorname{rank}BA=operatorname{rank}B$.
$endgroup$
– Bernard
Dec 12 '15 at 19:16
add a comment |
$begingroup$
Concerning the first point, remember $A$ and $B$ represent endomorphisms $f$ and $g$ in some basis, and $AB$ represent $fcirc g$.
Now $;DeclareMathOperatorrk{rank}DeclareMathOperatorimg{Im}rk B=1$ means $;dimimg g=1$. Then we know $;dimimg(fcirc g)ledimimg g$, which means $;rk ABlerk B$. So $rk AB=0$ or $1$.
For the second point, if $AB$ has indeed rank $1$, its kernel has dimension $n-1$. As the kernel is the eigenspace for the eigenvalue $0$, and has dimension at most equal to the multiplicity of this eigenvalues, it implies the eigenvalue $0$ has multiplicity at least $n-1$, and it can't have multiplicity $n$ as it would imply $rk AB=0$.
answered Dec 12 '15 at 14:38
BernardBernard
119k740113
$endgroup$
Concerning the first point, remember $A$ and $B$ represent endomorphisms $f$ and $g$ in some basis, and $AB$ represent $fcirc g$.
Now $;DeclareMathOperatorrk{rank}DeclareMathOperatorimg{Im}rk B=1$ means $;dimimg g=1$. Then we know $;dimimg(fcirc g)ledimimg g$, which means $;rk ABlerk B$. So $rk AB=0$ or $1$.
For the second point, if $AB$ has indeed rank $1$, its kernel has dimension $n-1$. As the kernel is the eigenspace for the eigenvalue $0$, and has dimension at most equal to the multiplicity of this eigenvalues, it implies the eigenvalue $0$ has multiplicity at least $n-1$, and it can't have multiplicity $n$ as it would imply $rk AB=0$.
answered Dec 12 '15 at 14:38
BernardBernard
119k740113
answered Dec 12 '15 at 14:38
BernardBernard
119k740113
answered Dec 12 '15 at 14:38
BernardBernard
119k740113
answered Dec 12 '15 at 14:38
BernardBernard
119k740113
119k740113
$begingroup$
Hi @Bernard, I apologize for probably withholding a key piece of information that I just found from rereading that solution. A is invertible. Does that now make $AB$ have rank 1, for certain? Thanks,
$endgroup$
– User001
Dec 12 '15 at 14:41
2
$begingroup$
Yes: if $A$ is invertible, $operatorname{rank}AB=operatorname{rank}B$, and similarly , $operatorname{rank}BA=operatorname{rank}B$.
$endgroup$
– Bernard
Dec 12 '15 at 19:16
add a comment |
$begingroup$
Hi @Bernard, I apologize for probably withholding a key piece of information that I just found from rereading that solution. A is invertible. Does that now make $AB$ have rank 1, for certain? Thanks,
$endgroup$
– User001
Dec 12 '15 at 14:41
2
$begingroup$
Yes: if $A$ is invertible, $operatorname{rank}AB=operatorname{rank}B$, and similarly , $operatorname{rank}BA=operatorname{rank}B$.
$endgroup$
– Bernard
Dec 12 '15 at 19:16
$begingroup$
Hi @Bernard, I apologize for probably withholding a key piece of information that I just found from rereading that solution. A is invertible. Does that now make $AB$ have rank 1, for certain? Thanks,
$endgroup$
– User001
Dec 12 '15 at 14:41
$begingroup$
Hi @Bernard, I apologize for probably withholding a key piece of information that I just found from rereading that solution. A is invertible. Does that now make $AB$ have rank 1, for certain? Thanks,
$endgroup$
– User001
Dec 12 '15 at 14:41
2
2
$begingroup$
Yes: if $A$ is invertible, $operatorname{rank}AB=operatorname{rank}B$, and similarly , $operatorname{rank}BA=operatorname{rank}B$.
$endgroup$
– Bernard
Dec 12 '15 at 19:16
$begingroup$
Yes: if $A$ is invertible, $operatorname{rank}AB=operatorname{rank}B$, and similarly , $operatorname{rank}BA=operatorname{rank}B$.
$endgroup$
– Bernard
Dec 12 '15 at 19:16
add a comment |
$begingroup$
The rank of a matrix is the dimension of its image ${rm rank} B = dim {Bx: x in mathbb R^n},$ or equivalently the dimension of the space spanned by its columns.
Since applying $AB$ on $mathbb R^n$ is the same as applying $A$ on the one-dimensional space $Bmathbb R^n$, we see that $AB$ can only have at most rank 1. Note that rank 0 is possible when the columns of $B$ are orthogonal to the rows of $A$.
We have the rank-nullity theorem $n = {rm rank} B + dim ker B$ (i.e. what doesn't contribute to the dimensions of the image must land in the kernel.
If ${rm rank} B = 1$, then the dimension of the kernel must be $n-1$, i.e. the eigenspace for the eigenvalue is $n-1$. We can deduce that there is also another nonzero eigenvalue since otherwise we'd have ${rm rank} AB = 0$.
answered Dec 12 '15 at 14:35
RolandRoland
4,80211331
$endgroup$
$begingroup$
Hi @Roland, the matrix $A$ is actually assumed to be invertible. I reread the solution and noticed that. Does the invertibility of $A$ make $AB$ have rank 1, for certain? Thanks so much,
$endgroup$
– User001
Dec 12 '15 at 14:58
1
$begingroup$
"We can deduce that there is also another nonzero eigenvalue since otherwise we'd have rank B = 0." This is not true.
$endgroup$
– Calle
Dec 12 '15 at 15:14
1
$begingroup$
@User001: Yes, because then $A^{-1}(AB) = B$ has rank 1, and so $AB$ must have rank at least 1.
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 15:23
$begingroup$
HI @IlmariKaronen, how does your equation imply that $AB$ must have rank at least 1? It seems like you are using the invertibility of $A^{-1}$, but I'm not sure how...thanks,
$endgroup$
– User001
Dec 12 '15 at 15:28
2
$begingroup$
@User001: By the same argument given by Roland above, the product of two matrices cannot have a higher rank than either of the matrices alone. Since we know that $A^{-1}(AB) = (A^{-1}A)B = B$ has rank 1, this means that both $A^{-1}$ and $AB$ must have rank at least 1. (Of course, for $A^{-1}$, we actually know more; namely that, as invertible matrices, both $A$ and $A^{-1}$ in fact have full rank. In fact, we can prove this simply by replacing $B$ in the previous argument with, say, the identity matrix. But for the proof that $AB$ has rank 1, we don't really need this extra information.)
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 16:04
add a comment |
$begingroup$
The rank of a matrix is the dimension of its image ${rm rank} B = dim {Bx: x in mathbb R^n},$ or equivalently the dimension of the space spanned by its columns.
Since applying $AB$ on $mathbb R^n$ is the same as applying $A$ on the one-dimensional space $Bmathbb R^n$, we see that $AB$ can only have at most rank 1. Note that rank 0 is possible when the columns of $B$ are orthogonal to the rows of $A$.
We have the rank-nullity theorem $n = {rm rank} B + dim ker B$ (i.e. what doesn't contribute to the dimensions of the image must land in the kernel.
If ${rm rank} B = 1$, then the dimension of the kernel must be $n-1$, i.e. the eigenspace for the eigenvalue is $n-1$. We can deduce that there is also another nonzero eigenvalue since otherwise we'd have ${rm rank} AB = 0$.
answered Dec 12 '15 at 14:35
RolandRoland
4,80211331
$endgroup$
$begingroup$
Hi @Roland, the matrix $A$ is actually assumed to be invertible. I reread the solution and noticed that. Does the invertibility of $A$ make $AB$ have rank 1, for certain? Thanks so much,
$endgroup$
– User001
Dec 12 '15 at 14:58
1
$begingroup$
"We can deduce that there is also another nonzero eigenvalue since otherwise we'd have rank B = 0." This is not true.
$endgroup$
– Calle
Dec 12 '15 at 15:14
1
$begingroup$
@User001: Yes, because then $A^{-1}(AB) = B$ has rank 1, and so $AB$ must have rank at least 1.
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 15:23
$begingroup$
HI @IlmariKaronen, how does your equation imply that $AB$ must have rank at least 1? It seems like you are using the invertibility of $A^{-1}$, but I'm not sure how...thanks,
$endgroup$
– User001
Dec 12 '15 at 15:28
2
$begingroup$
@User001: By the same argument given by Roland above, the product of two matrices cannot have a higher rank than either of the matrices alone. Since we know that $A^{-1}(AB) = (A^{-1}A)B = B$ has rank 1, this means that both $A^{-1}$ and $AB$ must have rank at least 1. (Of course, for $A^{-1}$, we actually know more; namely that, as invertible matrices, both $A$ and $A^{-1}$ in fact have full rank. In fact, we can prove this simply by replacing $B$ in the previous argument with, say, the identity matrix. But for the proof that $AB$ has rank 1, we don't really need this extra information.)
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 16:04
add a comment |
$begingroup$
The rank of a matrix is the dimension of its image ${rm rank} B = dim {Bx: x in mathbb R^n},$ or equivalently the dimension of the space spanned by its columns.
Since applying $AB$ on $mathbb R^n$ is the same as applying $A$ on the one-dimensional space $Bmathbb R^n$, we see that $AB$ can only have at most rank 1. Note that rank 0 is possible when the columns of $B$ are orthogonal to the rows of $A$.
We have the rank-nullity theorem $n = {rm rank} B + dim ker B$ (i.e. what doesn't contribute to the dimensions of the image must land in the kernel.
If ${rm rank} B = 1$, then the dimension of the kernel must be $n-1$, i.e. the eigenspace for the eigenvalue is $n-1$. We can deduce that there is also another nonzero eigenvalue since otherwise we'd have ${rm rank} AB = 0$.
answered Dec 12 '15 at 14:35
RolandRoland
4,80211331
$endgroup$
The rank of a matrix is the dimension of its image ${rm rank} B = dim {Bx: x in mathbb R^n},$ or equivalently the dimension of the space spanned by its columns.
Since applying $AB$ on $mathbb R^n$ is the same as applying $A$ on the one-dimensional space $Bmathbb R^n$, we see that $AB$ can only have at most rank 1. Note that rank 0 is possible when the columns of $B$ are orthogonal to the rows of $A$.
We have the rank-nullity theorem $n = {rm rank} B + dim ker B$ (i.e. what doesn't contribute to the dimensions of the image must land in the kernel.
If ${rm rank} B = 1$, then the dimension of the kernel must be $n-1$, i.e. the eigenspace for the eigenvalue is $n-1$. We can deduce that there is also another nonzero eigenvalue since otherwise we'd have ${rm rank} AB = 0$.
answered Dec 12 '15 at 14:35
RolandRoland
4,80211331
edited Dec 12 '15 at 16:40
answered Dec 12 '15 at 14:35
RolandRoland
4,80211331
answered Dec 12 '15 at 14:35
RolandRoland
4,80211331
answered Dec 12 '15 at 14:35
RolandRoland
4,80211331
4,80211331
$begingroup$
Hi @Roland, the matrix $A$ is actually assumed to be invertible. I reread the solution and noticed that. Does the invertibility of $A$ make $AB$ have rank 1, for certain? Thanks so much,
$endgroup$
– User001
Dec 12 '15 at 14:58
1
$begingroup$
"We can deduce that there is also another nonzero eigenvalue since otherwise we'd have rank B = 0." This is not true.
$endgroup$
– Calle
Dec 12 '15 at 15:14
1
$begingroup$
@User001: Yes, because then $A^{-1}(AB) = B$ has rank 1, and so $AB$ must have rank at least 1.
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 15:23
$begingroup$
HI @IlmariKaronen, how does your equation imply that $AB$ must have rank at least 1? It seems like you are using the invertibility of $A^{-1}$, but I'm not sure how...thanks,
$endgroup$
– User001
Dec 12 '15 at 15:28
2
$begingroup$
@User001: By the same argument given by Roland above, the product of two matrices cannot have a higher rank than either of the matrices alone. Since we know that $A^{-1}(AB) = (A^{-1}A)B = B$ has rank 1, this means that both $A^{-1}$ and $AB$ must have rank at least 1. (Of course, for $A^{-1}$, we actually know more; namely that, as invertible matrices, both $A$ and $A^{-1}$ in fact have full rank. In fact, we can prove this simply by replacing $B$ in the previous argument with, say, the identity matrix. But for the proof that $AB$ has rank 1, we don't really need this extra information.)
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 16:04
add a comment |
$begingroup$
Hi @Roland, the matrix $A$ is actually assumed to be invertible. I reread the solution and noticed that. Does the invertibility of $A$ make $AB$ have rank 1, for certain? Thanks so much,
$endgroup$
– User001
Dec 12 '15 at 14:58
1
$begingroup$
"We can deduce that there is also another nonzero eigenvalue since otherwise we'd have rank B = 0." This is not true.
$endgroup$
– Calle
Dec 12 '15 at 15:14
1
$begingroup$
@User001: Yes, because then $A^{-1}(AB) = B$ has rank 1, and so $AB$ must have rank at least 1.
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 15:23
$begingroup$
HI @IlmariKaronen, how does your equation imply that $AB$ must have rank at least 1? It seems like you are using the invertibility of $A^{-1}$, but I'm not sure how...thanks,
$endgroup$
– User001
Dec 12 '15 at 15:28
2
$begingroup$
@User001: By the same argument given by Roland above, the product of two matrices cannot have a higher rank than either of the matrices alone. Since we know that $A^{-1}(AB) = (A^{-1}A)B = B$ has rank 1, this means that both $A^{-1}$ and $AB$ must have rank at least 1. (Of course, for $A^{-1}$, we actually know more; namely that, as invertible matrices, both $A$ and $A^{-1}$ in fact have full rank. In fact, we can prove this simply by replacing $B$ in the previous argument with, say, the identity matrix. But for the proof that $AB$ has rank 1, we don't really need this extra information.)
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 16:04
$begingroup$
Hi @Roland, the matrix $A$ is actually assumed to be invertible. I reread the solution and noticed that. Does the invertibility of $A$ make $AB$ have rank 1, for certain? Thanks so much,
$endgroup$
– User001
Dec 12 '15 at 14:58
$begingroup$
Hi @Roland, the matrix $A$ is actually assumed to be invertible. I reread the solution and noticed that. Does the invertibility of $A$ make $AB$ have rank 1, for certain? Thanks so much,
$endgroup$
– User001
Dec 12 '15 at 14:58
1
1
$begingroup$
"We can deduce that there is also another nonzero eigenvalue since otherwise we'd have rank B = 0." This is not true.
$endgroup$
– Calle
Dec 12 '15 at 15:14
$begingroup$
"We can deduce that there is also another nonzero eigenvalue since otherwise we'd have rank B = 0." This is not true.
$endgroup$
– Calle
Dec 12 '15 at 15:14
1
1
$begingroup$
@User001: Yes, because then $A^{-1}(AB) = B$ has rank 1, and so $AB$ must have rank at least 1.
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 15:23
$begingroup$
@User001: Yes, because then $A^{-1}(AB) = B$ has rank 1, and so $AB$ must have rank at least 1.
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 15:23
$begingroup$
HI @IlmariKaronen, how does your equation imply that $AB$ must have rank at least 1? It seems like you are using the invertibility of $A^{-1}$, but I'm not sure how...thanks,
$endgroup$
– User001
Dec 12 '15 at 15:28
$begingroup$
HI @IlmariKaronen, how does your equation imply that $AB$ must have rank at least 1? It seems like you are using the invertibility of $A^{-1}$, but I'm not sure how...thanks,
$endgroup$
– User001
Dec 12 '15 at 15:28
2
2
$begingroup$
@User001: By the same argument given by Roland above, the product of two matrices cannot have a higher rank than either of the matrices alone. Since we know that $A^{-1}(AB) = (A^{-1}A)B = B$ has rank 1, this means that both $A^{-1}$ and $AB$ must have rank at least 1. (Of course, for $A^{-1}$, we actually know more; namely that, as invertible matrices, both $A$ and $A^{-1}$ in fact have full rank. In fact, we can prove this simply by replacing $B$ in the previous argument with, say, the identity matrix. But for the proof that $AB$ has rank 1, we don't really need this extra information.)
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 16:04
$begingroup$
@User001: By the same argument given by Roland above, the product of two matrices cannot have a higher rank than either of the matrices alone. Since we know that $A^{-1}(AB) = (A^{-1}A)B = B$ has rank 1, this means that both $A^{-1}$ and $AB$ must have rank at least 1. (Of course, for $A^{-1}$, we actually know more; namely that, as invertible matrices, both $A$ and $A^{-1}$ in fact have full rank. In fact, we can prove this simply by replacing $B$ in the previous argument with, say, the identity matrix. But for the proof that $AB$ has rank 1, we don't really need this extra information.)
$endgroup$
– Ilmari Karonen
Dec 12 '15 at 16:04
add a comment |
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