Weird condition for null space and range implying invertibility












1














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







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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
    Nov 18 '18 at 22:35
















1














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







share|cite|improve this question
























  • 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
    Nov 18 '18 at 22:35














1












1








1


1





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







share|cite|improve this question















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






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edited Nov 19 '18 at 16:49

























asked Nov 18 '18 at 22:16









AstlyDichrar

39618




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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
    Nov 18 '18 at 22:35


















  • 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
    Nov 18 '18 at 22:35
















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
Nov 18 '18 at 22:35




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
Nov 18 '18 at 22:35










1 Answer
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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.






share|cite|improve this answer





















  • That seems good, and way easier than what I did. Is my comment wrong, though?
    – AstlyDichrar
    Nov 19 '18 at 16:00












  • 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
    Nov 19 '18 at 16:12








  • 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
    Nov 19 '18 at 16:46













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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.






share|cite|improve this answer





















  • That seems good, and way easier than what I did. Is my comment wrong, though?
    – AstlyDichrar
    Nov 19 '18 at 16:00












  • 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
    Nov 19 '18 at 16:12








  • 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
    Nov 19 '18 at 16:46


















1














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.






share|cite|improve this answer





















  • That seems good, and way easier than what I did. Is my comment wrong, though?
    – AstlyDichrar
    Nov 19 '18 at 16:00












  • 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
    Nov 19 '18 at 16:12








  • 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
    Nov 19 '18 at 16:46
















1












1








1






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 19 '18 at 2:00









OtZman

814




814












  • That seems good, and way easier than what I did. Is my comment wrong, though?
    – AstlyDichrar
    Nov 19 '18 at 16:00












  • 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
    Nov 19 '18 at 16:12








  • 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
    Nov 19 '18 at 16:46




















  • That seems good, and way easier than what I did. Is my comment wrong, though?
    – AstlyDichrar
    Nov 19 '18 at 16:00












  • 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
    Nov 19 '18 at 16:12








  • 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
    Nov 19 '18 at 16:46


















That seems good, and way easier than what I did. Is my comment wrong, though?
– AstlyDichrar
Nov 19 '18 at 16:00






That seems good, and way easier than what I did. Is my comment wrong, though?
– AstlyDichrar
Nov 19 '18 at 16:00














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
Nov 19 '18 at 16:12






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
Nov 19 '18 at 16:12






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
Nov 19 '18 at 16:46






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
Nov 19 '18 at 16:46




















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Can't locate Autom4te/ChannelDefs.pm in @INC (when it definitely is there)

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up vote 1 down vote favorite I've been running into trouble running make for a build process that I know works on a 32-bit Ubuntu VM. I am running a 64-bit Ubuntu VM, and I have a feeling that the 64-bit may be the problem, but am not entirely sure. Basically, when I run the make command, I get the following error: Can't locate Autom4te/ChannelDefs.pm in @INC (@INC contains: [...]/staging_dir/host/share/autoconf /etc/perl /usr/local/lib/perl/5.14.2 /usr/local/share/perl/5.14.2 /usr/lib/perl5 /usr/share/perl5 /usr/lib/perl/5.14 /usr/share/perl/5.14 /usr/local/lib/site_perl .) at [...]/staging_dir/host/bin/autoreconf line 40. Now if I navigate to [...]/staging_dir/host/share/autoconf I can see that, contrary to what autoreconf thinks, Autom4te/ChannelDefs.pm definitely exists, so I don't really understand what
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