Mixing
Weird condition for null space and range implying invertibility
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The question is:
Let $A = begin{bmatrix} A_1 \ A_2end{bmatrix}in mathbb{M}_{ntimes n}(mathbb{C})$ (an $ntimes n$ matrix with entries on $mathbb{C}$) and suppose that $mathcal{N}(A_1)=mathcal{R}(A_2^top)$ ($mathcal{N}$ being the null space, and $mathcal{R}$ the range). Prove that $A$ is invertible.
My thought process was, to prove that $A$ is invertible, it seems reasonable that from what we're given I'm gonna try to prove that $n(A)=0$ (the dimension of the null space of $A$ is $0$).
Well, we have that $dim(mathcal{N}(A))=dim(mathcal{N}(A_1)capmathcal{N}(A_2))$, which is equal to $dim(mathcal{N}(A_1))+dim(mathcal{N}(A_2))-dim(mathcal{N}(A_1)+mathcal{N}(A_2))$. By the hypothesis, $mathcal{N}(A_1)=mathcal{R}(A_2^top)$, and $mathcal{R}(A_2^top)=mathcal{R}(A_2)$, so we're left with
$$dim(mathcal{N}(A))=n-dim(mathcal{N}(A_1)+mathcal{N}(A_2))$$
but I can't find a way to justify why $dim(mathcal{N}(A_1)+mathcal{N}(A_2))=n$, which is what it has to be if $A$ is invertible...
Edit: where $mathcal{R}(A_2^top)=mathcal{R}(A_2)$ is, it's actually meant to be $r(A_2)=r(A_2^top)$ (the rank of these matrices is equal).
linear-algebra
asked 2 days ago
AstlyDichrar
38118
add a comment |
up vote
1
down vote
favorite
The question is:
Let $A = begin{bmatrix} A_1 \ A_2end{bmatrix}in mathbb{M}_{ntimes n}(mathbb{C})$ (an $ntimes n$ matrix with entries on $mathbb{C}$) and suppose that $mathcal{N}(A_1)=mathcal{R}(A_2^top)$ ($mathcal{N}$ being the null space, and $mathcal{R}$ the range). Prove that $A$ is invertible.
My thought process was, to prove that $A$ is invertible, it seems reasonable that from what we're given I'm gonna try to prove that $n(A)=0$ (the dimension of the null space of $A$ is $0$).
Well, we have that $dim(mathcal{N}(A))=dim(mathcal{N}(A_1)capmathcal{N}(A_2))$, which is equal to $dim(mathcal{N}(A_1))+dim(mathcal{N}(A_2))-dim(mathcal{N}(A_1)+mathcal{N}(A_2))$. By the hypothesis, $mathcal{N}(A_1)=mathcal{R}(A_2^top)$, and $mathcal{R}(A_2^top)=mathcal{R}(A_2)$, so we're left with
$$dim(mathcal{N}(A))=n-dim(mathcal{N}(A_1)+mathcal{N}(A_2))$$
but I can't find a way to justify why $dim(mathcal{N}(A_1)+mathcal{N}(A_2))=n$, which is what it has to be if $A$ is invertible...
Edit: where $mathcal{R}(A_2^top)=mathcal{R}(A_2)$ is, it's actually meant to be $r(A_2)=r(A_2^top)$ (the rank of these matrices is equal).
linear-algebra
asked 2 days ago
AstlyDichrar
38118
I think I've got it... aren't $mathcal{R}(A_2^top)$ and $mathcal{N}(A_2)$ orthogonal subspaces? The dimension of that sum would then be $n$!
– AstlyDichrar
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The question is:
Let $A = begin{bmatrix} A_1 \ A_2end{bmatrix}in mathbb{M}_{ntimes n}(mathbb{C})$ (an $ntimes n$ matrix with entries on $mathbb{C}$) and suppose that $mathcal{N}(A_1)=mathcal{R}(A_2^top)$ ($mathcal{N}$ being the null space, and $mathcal{R}$ the range). Prove that $A$ is invertible.
My thought process was, to prove that $A$ is invertible, it seems reasonable that from what we're given I'm gonna try to prove that $n(A)=0$ (the dimension of the null space of $A$ is $0$).
Well, we have that $dim(mathcal{N}(A))=dim(mathcal{N}(A_1)capmathcal{N}(A_2))$, which is equal to $dim(mathcal{N}(A_1))+dim(mathcal{N}(A_2))-dim(mathcal{N}(A_1)+mathcal{N}(A_2))$. By the hypothesis, $mathcal{N}(A_1)=mathcal{R}(A_2^top)$, and $mathcal{R}(A_2^top)=mathcal{R}(A_2)$, so we're left with
$$dim(mathcal{N}(A))=n-dim(mathcal{N}(A_1)+mathcal{N}(A_2))$$
but I can't find a way to justify why $dim(mathcal{N}(A_1)+mathcal{N}(A_2))=n$, which is what it has to be if $A$ is invertible...
Edit: where $mathcal{R}(A_2^top)=mathcal{R}(A_2)$ is, it's actually meant to be $r(A_2)=r(A_2^top)$ (the rank of these matrices is equal).
linear-algebra
asked 2 days ago
AstlyDichrar
38118
The question is:
Let $A = begin{bmatrix} A_1 \ A_2end{bmatrix}in mathbb{M}_{ntimes n}(mathbb{C})$ (an $ntimes n$ matrix with entries on $mathbb{C}$) and suppose that $mathcal{N}(A_1)=mathcal{R}(A_2^top)$ ($mathcal{N}$ being the null space, and $mathcal{R}$ the range). Prove that $A$ is invertible.
My thought process was, to prove that $A$ is invertible, it seems reasonable that from what we're given I'm gonna try to prove that $n(A)=0$ (the dimension of the null space of $A$ is $0$).
Well, we have that $dim(mathcal{N}(A))=dim(mathcal{N}(A_1)capmathcal{N}(A_2))$, which is equal to $dim(mathcal{N}(A_1))+dim(mathcal{N}(A_2))-dim(mathcal{N}(A_1)+mathcal{N}(A_2))$. By the hypothesis, $mathcal{N}(A_1)=mathcal{R}(A_2^top)$, and $mathcal{R}(A_2^top)=mathcal{R}(A_2)$, so we're left with
$$dim(mathcal{N}(A))=n-dim(mathcal{N}(A_1)+mathcal{N}(A_2))$$
but I can't find a way to justify why $dim(mathcal{N}(A_1)+mathcal{N}(A_2))=n$, which is what it has to be if $A$ is invertible...
Edit: where $mathcal{R}(A_2^top)=mathcal{R}(A_2)$ is, it's actually meant to be $r(A_2)=r(A_2^top)$ (the rank of these matrices is equal).
linear-algebra
linear-algebra
asked 2 days ago
AstlyDichrar
38118
asked 2 days ago
AstlyDichrar
38118
edited yesterday
asked 2 days ago
AstlyDichrar
38118
asked 2 days ago
AstlyDichrar
38118
asked 2 days ago
AstlyDichrar
38118
38118
I think I've got it... aren't $mathcal{R}(A_2^top)$ and $mathcal{N}(A_2)$ orthogonal subspaces? The dimension of that sum would then be $n$!
– AstlyDichrar
2 days ago
add a comment |
I think I've got it... aren't $mathcal{R}(A_2^top)$ and $mathcal{N}(A_2)$ orthogonal subspaces? The dimension of that sum would then be $n$!
– AstlyDichrar
2 days ago
I think I've got it... aren't $mathcal{R}(A_2^top)$ and $mathcal{N}(A_2)$ orthogonal subspaces? The dimension of that sum would then be $n$!
– AstlyDichrar
2 days ago
I think I've got it... aren't $mathcal{R}(A_2^top)$ and $mathcal{N}(A_2)$ orthogonal subspaces? The dimension of that sum would then be $n$!
– AstlyDichrar
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Assume that $A$ is not invertible. Then there exists a nonzero vector $v in mathbb{C}^n$ such that
$$0 = Av = begin{bmatrix}A_1v \ A_2vend{bmatrix}.$$
This implies that $v in mathcal{N}(A_1)$ and $v in mathcal{N}(A_2)$. But then $v in mathcal{R}(A_2^top)$, so there exists a vector $u$ such that $v = A_2^top u$, and hence
$$0 neq v^top v = (A_2^top u)^top v = u^top A_2 v = u^top 0 = 0,$$
a contradiction. So the assumption cannot be true, and $A$ is therefore invertible.
answered 2 days ago
OtZman
564
That seems good, and way easier than what I did. Is my comment wrong, though?
– AstlyDichrar
yesterday
It's not immediately clear to me why $$text{dim}(mathcal{N}(A_1) cap mathcal{N}(A_2)) = text{dim}(mathcal{N}(A_1)) + text{dim}(mathcal{N}(A_2)) - text{dim}(mathcal{N}(A_1) + mathcal{N}(A_2)).$$ It could be that I'm missing something obvious here. Also, we will typically not have $mathcal{R}(A_2^top) = mathcal{R}(A_2)$; indeed, since $A_2$ is not square, the length of the vectors in each of those spaces will not be the same. Your comment on $mathcal{R}(A_2^top)$ being orthogonal to $mathcal{N}(A_2)$ is correct though.
– OtZman
yesterday
1
The formula is for the dimension of the sum of two subspaces, it's well known (check 2.43 in Axler's Linear Algebra Done Right, for example, or this: mathonline.wikidot.com/the-dimension-of-a-sum-of-subspaces). The second part was my mistake, it's not $mathcal{R}(A_2^top)=mathcal{R}(A_2)$ that I meant to say, but actually $r(A_2)=r(A_2^top)$ (the rank of $A_2$ and its transpose is equal).
– AstlyDichrar
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Assume that $A$ is not invertible. Then there exists a nonzero vector $v in mathbb{C}^n$ such that
$$0 = Av = begin{bmatrix}A_1v \ A_2vend{bmatrix}.$$
This implies that $v in mathcal{N}(A_1)$ and $v in mathcal{N}(A_2)$. But then $v in mathcal{R}(A_2^top)$, so there exists a vector $u$ such that $v = A_2^top u$, and hence
$$0 neq v^top v = (A_2^top u)^top v = u^top A_2 v = u^top 0 = 0,$$
a contradiction. So the assumption cannot be true, and $A$ is therefore invertible.
answered 2 days ago
OtZman
564
That seems good, and way easier than what I did. Is my comment wrong, though?
– AstlyDichrar
yesterday
It's not immediately clear to me why $$text{dim}(mathcal{N}(A_1) cap mathcal{N}(A_2)) = text{dim}(mathcal{N}(A_1)) + text{dim}(mathcal{N}(A_2)) - text{dim}(mathcal{N}(A_1) + mathcal{N}(A_2)).$$ It could be that I'm missing something obvious here. Also, we will typically not have $mathcal{R}(A_2^top) = mathcal{R}(A_2)$; indeed, since $A_2$ is not square, the length of the vectors in each of those spaces will not be the same. Your comment on $mathcal{R}(A_2^top)$ being orthogonal to $mathcal{N}(A_2)$ is correct though.
– OtZman
yesterday
1
The formula is for the dimension of the sum of two subspaces, it's well known (check 2.43 in Axler's Linear Algebra Done Right, for example, or this: mathonline.wikidot.com/the-dimension-of-a-sum-of-subspaces). The second part was my mistake, it's not $mathcal{R}(A_2^top)=mathcal{R}(A_2)$ that I meant to say, but actually $r(A_2)=r(A_2^top)$ (the rank of $A_2$ and its transpose is equal).
– AstlyDichrar
yesterday
add a comment |
up vote
1
down vote
accepted
Assume that $A$ is not invertible. Then there exists a nonzero vector $v in mathbb{C}^n$ such that
$$0 = Av = begin{bmatrix}A_1v \ A_2vend{bmatrix}.$$
This implies that $v in mathcal{N}(A_1)$ and $v in mathcal{N}(A_2)$. But then $v in mathcal{R}(A_2^top)$, so there exists a vector $u$ such that $v = A_2^top u$, and hence
$$0 neq v^top v = (A_2^top u)^top v = u^top A_2 v = u^top 0 = 0,$$
a contradiction. So the assumption cannot be true, and $A$ is therefore invertible.
answered 2 days ago
OtZman
564
That seems good, and way easier than what I did. Is my comment wrong, though?
– AstlyDichrar
yesterday
It's not immediately clear to me why $$text{dim}(mathcal{N}(A_1) cap mathcal{N}(A_2)) = text{dim}(mathcal{N}(A_1)) + text{dim}(mathcal{N}(A_2)) - text{dim}(mathcal{N}(A_1) + mathcal{N}(A_2)).$$ It could be that I'm missing something obvious here. Also, we will typically not have $mathcal{R}(A_2^top) = mathcal{R}(A_2)$; indeed, since $A_2$ is not square, the length of the vectors in each of those spaces will not be the same. Your comment on $mathcal{R}(A_2^top)$ being orthogonal to $mathcal{N}(A_2)$ is correct though.
– OtZman
yesterday
1
The formula is for the dimension of the sum of two subspaces, it's well known (check 2.43 in Axler's Linear Algebra Done Right, for example, or this: mathonline.wikidot.com/the-dimension-of-a-sum-of-subspaces). The second part was my mistake, it's not $mathcal{R}(A_2^top)=mathcal{R}(A_2)$ that I meant to say, but actually $r(A_2)=r(A_2^top)$ (the rank of $A_2$ and its transpose is equal).
– AstlyDichrar
yesterday
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Assume that $A$ is not invertible. Then there exists a nonzero vector $v in mathbb{C}^n$ such that
$$0 = Av = begin{bmatrix}A_1v \ A_2vend{bmatrix}.$$
This implies that $v in mathcal{N}(A_1)$ and $v in mathcal{N}(A_2)$. But then $v in mathcal{R}(A_2^top)$, so there exists a vector $u$ such that $v = A_2^top u$, and hence
$$0 neq v^top v = (A_2^top u)^top v = u^top A_2 v = u^top 0 = 0,$$
a contradiction. So the assumption cannot be true, and $A$ is therefore invertible.
answered 2 days ago
OtZman
564
Assume that $A$ is not invertible. Then there exists a nonzero vector $v in mathbb{C}^n$ such that
$$0 = Av = begin{bmatrix}A_1v \ A_2vend{bmatrix}.$$
This implies that $v in mathcal{N}(A_1)$ and $v in mathcal{N}(A_2)$. But then $v in mathcal{R}(A_2^top)$, so there exists a vector $u$ such that $v = A_2^top u$, and hence
$$0 neq v^top v = (A_2^top u)^top v = u^top A_2 v = u^top 0 = 0,$$
a contradiction. So the assumption cannot be true, and $A$ is therefore invertible.
answered 2 days ago
OtZman
564
answered 2 days ago
OtZman
564
answered 2 days ago
OtZman
564
answered 2 days ago
OtZman
564
564
That seems good, and way easier than what I did. Is my comment wrong, though?
– AstlyDichrar
yesterday
It's not immediately clear to me why $$text{dim}(mathcal{N}(A_1) cap mathcal{N}(A_2)) = text{dim}(mathcal{N}(A_1)) + text{dim}(mathcal{N}(A_2)) - text{dim}(mathcal{N}(A_1) + mathcal{N}(A_2)).$$ It could be that I'm missing something obvious here. Also, we will typically not have $mathcal{R}(A_2^top) = mathcal{R}(A_2)$; indeed, since $A_2$ is not square, the length of the vectors in each of those spaces will not be the same. Your comment on $mathcal{R}(A_2^top)$ being orthogonal to $mathcal{N}(A_2)$ is correct though.
– OtZman
yesterday
1
The formula is for the dimension of the sum of two subspaces, it's well known (check 2.43 in Axler's Linear Algebra Done Right, for example, or this: mathonline.wikidot.com/the-dimension-of-a-sum-of-subspaces). The second part was my mistake, it's not $mathcal{R}(A_2^top)=mathcal{R}(A_2)$ that I meant to say, but actually $r(A_2)=r(A_2^top)$ (the rank of $A_2$ and its transpose is equal).
– AstlyDichrar
yesterday
add a comment |
That seems good, and way easier than what I did. Is my comment wrong, though?
– AstlyDichrar
yesterday
It's not immediately clear to me why $$text{dim}(mathcal{N}(A_1) cap mathcal{N}(A_2)) = text{dim}(mathcal{N}(A_1)) + text{dim}(mathcal{N}(A_2)) - text{dim}(mathcal{N}(A_1) + mathcal{N}(A_2)).$$ It could be that I'm missing something obvious here. Also, we will typically not have $mathcal{R}(A_2^top) = mathcal{R}(A_2)$; indeed, since $A_2$ is not square, the length of the vectors in each of those spaces will not be the same. Your comment on $mathcal{R}(A_2^top)$ being orthogonal to $mathcal{N}(A_2)$ is correct though.
– OtZman
yesterday
1
The formula is for the dimension of the sum of two subspaces, it's well known (check 2.43 in Axler's Linear Algebra Done Right, for example, or this: mathonline.wikidot.com/the-dimension-of-a-sum-of-subspaces). The second part was my mistake, it's not $mathcal{R}(A_2^top)=mathcal{R}(A_2)$ that I meant to say, but actually $r(A_2)=r(A_2^top)$ (the rank of $A_2$ and its transpose is equal).
– AstlyDichrar
yesterday
That seems good, and way easier than what I did. Is my comment wrong, though?
– AstlyDichrar
yesterday
That seems good, and way easier than what I did. Is my comment wrong, though?
– AstlyDichrar
yesterday
It's not immediately clear to me why $$text{dim}(mathcal{N}(A_1) cap mathcal{N}(A_2)) = text{dim}(mathcal{N}(A_1)) + text{dim}(mathcal{N}(A_2)) - text{dim}(mathcal{N}(A_1) + mathcal{N}(A_2)).$$ It could be that I'm missing something obvious here. Also, we will typically not have $mathcal{R}(A_2^top) = mathcal{R}(A_2)$; indeed, since $A_2$ is not square, the length of the vectors in each of those spaces will not be the same. Your comment on $mathcal{R}(A_2^top)$ being orthogonal to $mathcal{N}(A_2)$ is correct though.
– OtZman
yesterday
It's not immediately clear to me why $$text{dim}(mathcal{N}(A_1) cap mathcal{N}(A_2)) = text{dim}(mathcal{N}(A_1)) + text{dim}(mathcal{N}(A_2)) - text{dim}(mathcal{N}(A_1) + mathcal{N}(A_2)).$$ It could be that I'm missing something obvious here. Also, we will typically not have $mathcal{R}(A_2^top) = mathcal{R}(A_2)$; indeed, since $A_2$ is not square, the length of the vectors in each of those spaces will not be the same. Your comment on $mathcal{R}(A_2^top)$ being orthogonal to $mathcal{N}(A_2)$ is correct though.
– OtZman
yesterday
1
1
The formula is for the dimension of the sum of two subspaces, it's well known (check 2.43 in Axler's Linear Algebra Done Right, for example, or this: mathonline.wikidot.com/the-dimension-of-a-sum-of-subspaces). The second part was my mistake, it's not $mathcal{R}(A_2^top)=mathcal{R}(A_2)$ that I meant to say, but actually $r(A_2)=r(A_2^top)$ (the rank of $A_2$ and its transpose is equal).
– AstlyDichrar
yesterday
The formula is for the dimension of the sum of two subspaces, it's well known (check 2.43 in Axler's Linear Algebra Done Right, for example, or this: mathonline.wikidot.com/the-dimension-of-a-sum-of-subspaces). The second part was my mistake, it's not $mathcal{R}(A_2^top)=mathcal{R}(A_2)$ that I meant to say, but actually $r(A_2)=r(A_2^top)$ (the rank of $A_2$ and its transpose is equal).
– AstlyDichrar
yesterday
add a comment |
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I think I've got it... aren't $mathcal{R}(A_2^top)$ and $mathcal{N}(A_2)$ orthogonal subspaces? The dimension of that sum would then be $n$!
– AstlyDichrar
2 days ago