Mixing
Proof of $operatorname{Ran} A_e^T = operatorname{Ran} A^T$, where $operatorname{Ran}A^T$ is the range of...
$begingroup$
$newcommand{Ran}{operatorname{Ran}}$I am having trouble in understanding the following proof of $Ran A_e^T = Ran A^T$, where $A_e^T$ is echelon form of $A^T$, given in my book.
Proof :-
By the definition of echelon form, $A_e = E_{m times m} A_{mtimes n}$, where $E$ is the product of elementary matrices.
$$Ran(A_e^T) = Ran(A^T E^T) = A^T(Ran(E^T)) = A^T(Bbb R^m) = Ran(A^T)tag*{$Box$}$$
I know that for a linear tansfromation $T$ and a set of vectors $X$, $$T(X) := {mathbf y = T(mathbf x) : mathbf x in X }$$.
We can identity $A^T$ with the linear transformation $A^T : Bbb R^m to Bbb R^n$, so by definition $A^T(Bbb R^m) = Ran(A^T)$. Hence the last equality.
For the second equality from last, we use the fact that $E^T$ is invertible and therefore it is onto.
Is my reasoning correct for last and second from last equalities?
How did the author get second equality, $Ran(A^T E^T) = A^T(Ran(E^T))$ ? I don't think $Ran(AB) = A (Ran(B))$ is true.
linear-algebra matrices vector-spaces linear-transformations
edited Jan 31 at 17:52
Martin Sleziak
45k10122277
asked Oct 25 '17 at 18:57
user8277998user8277998
1,691521
$endgroup$
add a comment |
$begingroup$
$newcommand{Ran}{operatorname{Ran}}$I am having trouble in understanding the following proof of $Ran A_e^T = Ran A^T$, where $A_e^T$ is echelon form of $A^T$, given in my book.
Proof :-
By the definition of echelon form, $A_e = E_{m times m} A_{mtimes n}$, where $E$ is the product of elementary matrices.
$$Ran(A_e^T) = Ran(A^T E^T) = A^T(Ran(E^T)) = A^T(Bbb R^m) = Ran(A^T)tag*{$Box$}$$
I know that for a linear tansfromation $T$ and a set of vectors $X$, $$T(X) := {mathbf y = T(mathbf x) : mathbf x in X }$$.
We can identity $A^T$ with the linear transformation $A^T : Bbb R^m to Bbb R^n$, so by definition $A^T(Bbb R^m) = Ran(A^T)$. Hence the last equality.
For the second equality from last, we use the fact that $E^T$ is invertible and therefore it is onto.
Is my reasoning correct for last and second from last equalities?
How did the author get second equality, $Ran(A^T E^T) = A^T(Ran(E^T))$ ? I don't think $Ran(AB) = A (Ran(B))$ is true.
linear-algebra matrices vector-spaces linear-transformations
edited Jan 31 at 17:52
Martin Sleziak
45k10122277
asked Oct 25 '17 at 18:57
user8277998user8277998
1,691521
$endgroup$
$begingroup$
Does $text{Ran}(A)$ mean rank of $A$?
$endgroup$
– Tengu
Oct 26 '17 at 20:32
$begingroup$
@ToanPham No. It means column space of $A$ or range of $A$.
$endgroup$
– user8277998
Oct 27 '17 at 8:51
add a comment |
$begingroup$
$newcommand{Ran}{operatorname{Ran}}$I am having trouble in understanding the following proof of $Ran A_e^T = Ran A^T$, where $A_e^T$ is echelon form of $A^T$, given in my book.
Proof :-
By the definition of echelon form, $A_e = E_{m times m} A_{mtimes n}$, where $E$ is the product of elementary matrices.
$$Ran(A_e^T) = Ran(A^T E^T) = A^T(Ran(E^T)) = A^T(Bbb R^m) = Ran(A^T)tag*{$Box$}$$
I know that for a linear tansfromation $T$ and a set of vectors $X$, $$T(X) := {mathbf y = T(mathbf x) : mathbf x in X }$$.
We can identity $A^T$ with the linear transformation $A^T : Bbb R^m to Bbb R^n$, so by definition $A^T(Bbb R^m) = Ran(A^T)$. Hence the last equality.
For the second equality from last, we use the fact that $E^T$ is invertible and therefore it is onto.
Is my reasoning correct for last and second from last equalities?
How did the author get second equality, $Ran(A^T E^T) = A^T(Ran(E^T))$ ? I don't think $Ran(AB) = A (Ran(B))$ is true.
linear-algebra matrices vector-spaces linear-transformations
edited Jan 31 at 17:52
Martin Sleziak
45k10122277
asked Oct 25 '17 at 18:57
user8277998user8277998
1,691521
$endgroup$
$newcommand{Ran}{operatorname{Ran}}$I am having trouble in understanding the following proof of $Ran A_e^T = Ran A^T$, where $A_e^T$ is echelon form of $A^T$, given in my book.
Proof :-
By the definition of echelon form, $A_e = E_{m times m} A_{mtimes n}$, where $E$ is the product of elementary matrices.
$$Ran(A_e^T) = Ran(A^T E^T) = A^T(Ran(E^T)) = A^T(Bbb R^m) = Ran(A^T)tag*{$Box$}$$
I know that for a linear tansfromation $T$ and a set of vectors $X$, $$T(X) := {mathbf y = T(mathbf x) : mathbf x in X }$$.
We can identity $A^T$ with the linear transformation $A^T : Bbb R^m to Bbb R^n$, so by definition $A^T(Bbb R^m) = Ran(A^T)$. Hence the last equality.
For the second equality from last, we use the fact that $E^T$ is invertible and therefore it is onto.
Is my reasoning correct for last and second from last equalities?
How did the author get second equality, $Ran(A^T E^T) = A^T(Ran(E^T))$ ? I don't think $Ran(AB) = A (Ran(B))$ is true.
linear-algebra matrices vector-spaces linear-transformations
linear-algebra matrices vector-spaces linear-transformations
edited Jan 31 at 17:52
Martin Sleziak
45k10122277
asked Oct 25 '17 at 18:57
user8277998user8277998
1,691521
edited Jan 31 at 17:52
Martin Sleziak
45k10122277
asked Oct 25 '17 at 18:57
user8277998user8277998
1,691521
edited Jan 31 at 17:52
Martin Sleziak
45k10122277
edited Jan 31 at 17:52
Martin Sleziak
45k10122277
edited Jan 31 at 17:52
Martin Sleziak
45k10122277
45k10122277
asked Oct 25 '17 at 18:57
user8277998user8277998
1,691521
asked Oct 25 '17 at 18:57
user8277998user8277998
1,691521
asked Oct 25 '17 at 18:57
user8277998user8277998
1,691521
1,691521
$begingroup$
Does $text{Ran}(A)$ mean rank of $A$?
$endgroup$
– Tengu
Oct 26 '17 at 20:32
$begingroup$
@ToanPham No. It means column space of $A$ or range of $A$.
$endgroup$
– user8277998
Oct 27 '17 at 8:51
add a comment |
$begingroup$
Does $text{Ran}(A)$ mean rank of $A$?
$endgroup$
– Tengu
Oct 26 '17 at 20:32
$begingroup$
@ToanPham No. It means column space of $A$ or range of $A$.
$endgroup$
– user8277998
Oct 27 '17 at 8:51
$begingroup$
Does $text{Ran}(A)$ mean rank of $A$?
$endgroup$
– Tengu
Oct 26 '17 at 20:32
$begingroup$
Does $text{Ran}(A)$ mean rank of $A$?
$endgroup$
– Tengu
Oct 26 '17 at 20:32
$begingroup$
@ToanPham No. It means column space of $A$ or range of $A$.
$endgroup$
– user8277998
Oct 27 '17 at 8:51
$begingroup$
@ToanPham No. It means column space of $A$ or range of $A$.
$endgroup$
– user8277998
Oct 27 '17 at 8:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $text{Ran }A_e^T=text{Ran }A^T$ then I believe $A_e^T$ is not the echelon form of $A^T$, i.e. $(A^T)_e$ but is rather the transpose of $A_e$, i.e. $A_e^T=(A_e)^T$. This explains why we can have $A_e^T=(EA)^T=A^TE^T$.
Show $text{Ran}(A^TE^T)=A^Ttext{Ran }(E^T)$.
Note that any $i$-th column vector in $A^TE^T$ equals $A^T cdot mathbf{c}_i$ where $mathbf{c_i}$ is the column vector of $E^T$. Every vector in the column space of $A^TE^T$ can be written as a linear combination of column vectors of $A^TE^T$, which then follows that they can be written as $A^T$ times a vector in column space of $E^T$. This results $text{Ran }(A^TE^T)=A^T text{Ran }(E^T)$.
Since $E$ is invertible so $text{Ran } E^T=mathbf{R}^m$.
Show $A^T(mathbf{R}^m)=text{Ran }(A^T)$.
Similar to previous argument, every vector in column space of $A^T$ can be written as $A^T$ times a vector in $mathbf{R}^m$.
answered Oct 28 '17 at 13:57
TenguTengu
2,65411021
$endgroup$
$begingroup$
So $Ran AB = ARan B$ true in general ? BTW you don't need to usetext{Ran }on this page, you can useRan.
$endgroup$
– user8277998
Oct 28 '17 at 21:01
$begingroup$
Also when we write $A^T Bbb R^m$ what does it mean, we can't just multiply a martix and a vector space.
$endgroup$
– user8277998
Oct 28 '17 at 21:13
$begingroup$
Thanks for the tip. Yes, $Ran AB=A Ran B$ is true in general. And for your second question: This is the first time I saw the notation $Ran$ so I'm not sure, but according to the above, we don't multiply a matrix to a vector space but rather a matrix to every vector (also matrix) in a vector space so we actually can do that (given certain conditions occur).
$endgroup$
– Tengu
Oct 28 '17 at 22:20
$begingroup$
Since $Ran AB = ARan B$ is true then is $Ran A^T I = A^T Ran I = A^T Bbb R^m$ correct proof for $A^T Bbb R^m = Ran A^T$ ?
$endgroup$
– user8277998
Oct 29 '17 at 11:16
$begingroup$
Sorry I am still confused about how we can do $A^T Bbb R^m$, If we consider $Bbb R^m$ as a matrix of all vectors in $Bbb R^m$ then it is a infinite matrix.
$endgroup$
– user8277998
Oct 29 '17 at 11:23
|
show 3 more comments
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1 Answer
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$begingroup$
If $text{Ran }A_e^T=text{Ran }A^T$ then I believe $A_e^T$ is not the echelon form of $A^T$, i.e. $(A^T)_e$ but is rather the transpose of $A_e$, i.e. $A_e^T=(A_e)^T$. This explains why we can have $A_e^T=(EA)^T=A^TE^T$.
Show $text{Ran}(A^TE^T)=A^Ttext{Ran }(E^T)$.
Note that any $i$-th column vector in $A^TE^T$ equals $A^T cdot mathbf{c}_i$ where $mathbf{c_i}$ is the column vector of $E^T$. Every vector in the column space of $A^TE^T$ can be written as a linear combination of column vectors of $A^TE^T$, which then follows that they can be written as $A^T$ times a vector in column space of $E^T$. This results $text{Ran }(A^TE^T)=A^T text{Ran }(E^T)$.
Since $E$ is invertible so $text{Ran } E^T=mathbf{R}^m$.
Show $A^T(mathbf{R}^m)=text{Ran }(A^T)$.
Similar to previous argument, every vector in column space of $A^T$ can be written as $A^T$ times a vector in $mathbf{R}^m$.
answered Oct 28 '17 at 13:57
TenguTengu
2,65411021
$endgroup$
$begingroup$
So $Ran AB = ARan B$ true in general ? BTW you don't need to usetext{Ran }on this page, you can useRan.
$endgroup$
– user8277998
Oct 28 '17 at 21:01
$begingroup$
Also when we write $A^T Bbb R^m$ what does it mean, we can't just multiply a martix and a vector space.
$endgroup$
– user8277998
Oct 28 '17 at 21:13
$begingroup$
Thanks for the tip. Yes, $Ran AB=A Ran B$ is true in general. And for your second question: This is the first time I saw the notation $Ran$ so I'm not sure, but according to the above, we don't multiply a matrix to a vector space but rather a matrix to every vector (also matrix) in a vector space so we actually can do that (given certain conditions occur).
$endgroup$
– Tengu
Oct 28 '17 at 22:20
$begingroup$
Since $Ran AB = ARan B$ is true then is $Ran A^T I = A^T Ran I = A^T Bbb R^m$ correct proof for $A^T Bbb R^m = Ran A^T$ ?
$endgroup$
– user8277998
Oct 29 '17 at 11:16
$begingroup$
Sorry I am still confused about how we can do $A^T Bbb R^m$, If we consider $Bbb R^m$ as a matrix of all vectors in $Bbb R^m$ then it is a infinite matrix.
$endgroup$
– user8277998
Oct 29 '17 at 11:23
|
show 3 more comments
$begingroup$
If $text{Ran }A_e^T=text{Ran }A^T$ then I believe $A_e^T$ is not the echelon form of $A^T$, i.e. $(A^T)_e$ but is rather the transpose of $A_e$, i.e. $A_e^T=(A_e)^T$. This explains why we can have $A_e^T=(EA)^T=A^TE^T$.
Show $text{Ran}(A^TE^T)=A^Ttext{Ran }(E^T)$.
Note that any $i$-th column vector in $A^TE^T$ equals $A^T cdot mathbf{c}_i$ where $mathbf{c_i}$ is the column vector of $E^T$. Every vector in the column space of $A^TE^T$ can be written as a linear combination of column vectors of $A^TE^T$, which then follows that they can be written as $A^T$ times a vector in column space of $E^T$. This results $text{Ran }(A^TE^T)=A^T text{Ran }(E^T)$.
Since $E$ is invertible so $text{Ran } E^T=mathbf{R}^m$.
Show $A^T(mathbf{R}^m)=text{Ran }(A^T)$.
Similar to previous argument, every vector in column space of $A^T$ can be written as $A^T$ times a vector in $mathbf{R}^m$.
answered Oct 28 '17 at 13:57
TenguTengu
2,65411021
$endgroup$
$begingroup$
So $Ran AB = ARan B$ true in general ? BTW you don't need to usetext{Ran }on this page, you can useRan.
$endgroup$
– user8277998
Oct 28 '17 at 21:01
$begingroup$
Also when we write $A^T Bbb R^m$ what does it mean, we can't just multiply a martix and a vector space.
$endgroup$
– user8277998
Oct 28 '17 at 21:13
$begingroup$
Thanks for the tip. Yes, $Ran AB=A Ran B$ is true in general. And for your second question: This is the first time I saw the notation $Ran$ so I'm not sure, but according to the above, we don't multiply a matrix to a vector space but rather a matrix to every vector (also matrix) in a vector space so we actually can do that (given certain conditions occur).
$endgroup$
– Tengu
Oct 28 '17 at 22:20
$begingroup$
Since $Ran AB = ARan B$ is true then is $Ran A^T I = A^T Ran I = A^T Bbb R^m$ correct proof for $A^T Bbb R^m = Ran A^T$ ?
$endgroup$
– user8277998
Oct 29 '17 at 11:16
$begingroup$
Sorry I am still confused about how we can do $A^T Bbb R^m$, If we consider $Bbb R^m$ as a matrix of all vectors in $Bbb R^m$ then it is a infinite matrix.
$endgroup$
– user8277998
Oct 29 '17 at 11:23
|
show 3 more comments
$begingroup$
If $text{Ran }A_e^T=text{Ran }A^T$ then I believe $A_e^T$ is not the echelon form of $A^T$, i.e. $(A^T)_e$ but is rather the transpose of $A_e$, i.e. $A_e^T=(A_e)^T$. This explains why we can have $A_e^T=(EA)^T=A^TE^T$.
Show $text{Ran}(A^TE^T)=A^Ttext{Ran }(E^T)$.
Note that any $i$-th column vector in $A^TE^T$ equals $A^T cdot mathbf{c}_i$ where $mathbf{c_i}$ is the column vector of $E^T$. Every vector in the column space of $A^TE^T$ can be written as a linear combination of column vectors of $A^TE^T$, which then follows that they can be written as $A^T$ times a vector in column space of $E^T$. This results $text{Ran }(A^TE^T)=A^T text{Ran }(E^T)$.
Since $E$ is invertible so $text{Ran } E^T=mathbf{R}^m$.
Show $A^T(mathbf{R}^m)=text{Ran }(A^T)$.
Similar to previous argument, every vector in column space of $A^T$ can be written as $A^T$ times a vector in $mathbf{R}^m$.
answered Oct 28 '17 at 13:57
TenguTengu
2,65411021
$endgroup$
If $text{Ran }A_e^T=text{Ran }A^T$ then I believe $A_e^T$ is not the echelon form of $A^T$, i.e. $(A^T)_e$ but is rather the transpose of $A_e$, i.e. $A_e^T=(A_e)^T$. This explains why we can have $A_e^T=(EA)^T=A^TE^T$.
Show $text{Ran}(A^TE^T)=A^Ttext{Ran }(E^T)$.
Note that any $i$-th column vector in $A^TE^T$ equals $A^T cdot mathbf{c}_i$ where $mathbf{c_i}$ is the column vector of $E^T$. Every vector in the column space of $A^TE^T$ can be written as a linear combination of column vectors of $A^TE^T$, which then follows that they can be written as $A^T$ times a vector in column space of $E^T$. This results $text{Ran }(A^TE^T)=A^T text{Ran }(E^T)$.
Since $E$ is invertible so $text{Ran } E^T=mathbf{R}^m$.
Show $A^T(mathbf{R}^m)=text{Ran }(A^T)$.
Similar to previous argument, every vector in column space of $A^T$ can be written as $A^T$ times a vector in $mathbf{R}^m$.
answered Oct 28 '17 at 13:57
TenguTengu
2,65411021
answered Oct 28 '17 at 13:57
TenguTengu
2,65411021
answered Oct 28 '17 at 13:57
TenguTengu
2,65411021
answered Oct 28 '17 at 13:57
TenguTengu
2,65411021
2,65411021
$begingroup$
So $Ran AB = ARan B$ true in general ? BTW you don't need to usetext{Ran }on this page, you can useRan.
$endgroup$
– user8277998
Oct 28 '17 at 21:01
$begingroup$
Also when we write $A^T Bbb R^m$ what does it mean, we can't just multiply a martix and a vector space.
$endgroup$
– user8277998
Oct 28 '17 at 21:13
$begingroup$
Thanks for the tip. Yes, $Ran AB=A Ran B$ is true in general. And for your second question: This is the first time I saw the notation $Ran$ so I'm not sure, but according to the above, we don't multiply a matrix to a vector space but rather a matrix to every vector (also matrix) in a vector space so we actually can do that (given certain conditions occur).
$endgroup$
– Tengu
Oct 28 '17 at 22:20
$begingroup$
Since $Ran AB = ARan B$ is true then is $Ran A^T I = A^T Ran I = A^T Bbb R^m$ correct proof for $A^T Bbb R^m = Ran A^T$ ?
$endgroup$
– user8277998
Oct 29 '17 at 11:16
$begingroup$
Sorry I am still confused about how we can do $A^T Bbb R^m$, If we consider $Bbb R^m$ as a matrix of all vectors in $Bbb R^m$ then it is a infinite matrix.
$endgroup$
– user8277998
Oct 29 '17 at 11:23
|
show 3 more comments
$begingroup$
So $Ran AB = ARan B$ true in general ? BTW you don't need to usetext{Ran }on this page, you can useRan.
$endgroup$
– user8277998
Oct 28 '17 at 21:01
$begingroup$
Also when we write $A^T Bbb R^m$ what does it mean, we can't just multiply a martix and a vector space.
$endgroup$
– user8277998
Oct 28 '17 at 21:13
$begingroup$
Thanks for the tip. Yes, $Ran AB=A Ran B$ is true in general. And for your second question: This is the first time I saw the notation $Ran$ so I'm not sure, but according to the above, we don't multiply a matrix to a vector space but rather a matrix to every vector (also matrix) in a vector space so we actually can do that (given certain conditions occur).
$endgroup$
– Tengu
Oct 28 '17 at 22:20
$begingroup$
Since $Ran AB = ARan B$ is true then is $Ran A^T I = A^T Ran I = A^T Bbb R^m$ correct proof for $A^T Bbb R^m = Ran A^T$ ?
$endgroup$
– user8277998
Oct 29 '17 at 11:16
$begingroup$
Sorry I am still confused about how we can do $A^T Bbb R^m$, If we consider $Bbb R^m$ as a matrix of all vectors in $Bbb R^m$ then it is a infinite matrix.
$endgroup$
– user8277998
Oct 29 '17 at 11:23
$begingroup$
So $Ran AB = ARan B$ true in general ? BTW you don't need to use
text{Ran } on this page, you can use Ran.$endgroup$
– user8277998
Oct 28 '17 at 21:01
$begingroup$
So $Ran AB = ARan B$ true in general ? BTW you don't need to use
text{Ran } on this page, you can use Ran.$endgroup$
– user8277998
Oct 28 '17 at 21:01
$begingroup$
Also when we write $A^T Bbb R^m$ what does it mean, we can't just multiply a martix and a vector space.
$endgroup$
– user8277998
Oct 28 '17 at 21:13
$begingroup$
Also when we write $A^T Bbb R^m$ what does it mean, we can't just multiply a martix and a vector space.
$endgroup$
– user8277998
Oct 28 '17 at 21:13
$begingroup$
Thanks for the tip. Yes, $Ran AB=A Ran B$ is true in general. And for your second question: This is the first time I saw the notation $Ran$ so I'm not sure, but according to the above, we don't multiply a matrix to a vector space but rather a matrix to every vector (also matrix) in a vector space so we actually can do that (given certain conditions occur).
$endgroup$
– Tengu
Oct 28 '17 at 22:20
$begingroup$
Thanks for the tip. Yes, $Ran AB=A Ran B$ is true in general. And for your second question: This is the first time I saw the notation $Ran$ so I'm not sure, but according to the above, we don't multiply a matrix to a vector space but rather a matrix to every vector (also matrix) in a vector space so we actually can do that (given certain conditions occur).
$endgroup$
– Tengu
Oct 28 '17 at 22:20
$begingroup$
Since $Ran AB = ARan B$ is true then is $Ran A^T I = A^T Ran I = A^T Bbb R^m$ correct proof for $A^T Bbb R^m = Ran A^T$ ?
$endgroup$
– user8277998
Oct 29 '17 at 11:16
$begingroup$
Since $Ran AB = ARan B$ is true then is $Ran A^T I = A^T Ran I = A^T Bbb R^m$ correct proof for $A^T Bbb R^m = Ran A^T$ ?
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– user8277998
Oct 29 '17 at 11:16
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Sorry I am still confused about how we can do $A^T Bbb R^m$, If we consider $Bbb R^m$ as a matrix of all vectors in $Bbb R^m$ then it is a infinite matrix.
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– user8277998
Oct 29 '17 at 11:23
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Sorry I am still confused about how we can do $A^T Bbb R^m$, If we consider $Bbb R^m$ as a matrix of all vectors in $Bbb R^m$ then it is a infinite matrix.
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– user8277998
Oct 29 '17 at 11:23
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Does $text{Ran}(A)$ mean rank of $A$?
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– Tengu
Oct 26 '17 at 20:32
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@ToanPham No. It means column space of $A$ or range of $A$.
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– user8277998
Oct 27 '17 at 8:51