Mixing
Proving the existance of a one- or two dimensional subspace that is both A and B invariant.
$begingroup$
I've just started learning about invariant subspaces and I came across this exercise:
Let $V$ be a finite-dimensional space over the field $mathbb{C}$. Prove that if endomorphism matrices $A$ and $B$ satisfy the equation $A^2=B^2=I$, there exists a one- or two-dimensional subspace that is both $A$ and $B$ invariant.
The only thing that I've proven thus far is that the only possible eigenvalues of $A$ and $B$ may be $1$ or $-1$ (If $v$ is an eigenvector of $A$ then $Av=lambda v Rightarrow Iv=A^2v=lambda Av=lambda^{2}v$ ; this can only happen if $lambda =pm 1$).
Any help is appreciated. Thanks.
matrices invariance
asked Jan 7 at 15:56
The CatThe Cat
18411
$endgroup$
add a comment |
$begingroup$
I've just started learning about invariant subspaces and I came across this exercise:
Let $V$ be a finite-dimensional space over the field $mathbb{C}$. Prove that if endomorphism matrices $A$ and $B$ satisfy the equation $A^2=B^2=I$, there exists a one- or two-dimensional subspace that is both $A$ and $B$ invariant.
The only thing that I've proven thus far is that the only possible eigenvalues of $A$ and $B$ may be $1$ or $-1$ (If $v$ is an eigenvector of $A$ then $Av=lambda v Rightarrow Iv=A^2v=lambda Av=lambda^{2}v$ ; this can only happen if $lambda =pm 1$).
Any help is appreciated. Thanks.
matrices invariance
asked Jan 7 at 15:56
The CatThe Cat
18411
$endgroup$
1
$begingroup$
you need to use that an invariant subspace allows a lot more than just eigenvectors (i.e. as soon as you find an honest eigenvector $v$ for $A $ you can pimp that one into an invariant subspace (since then $langle v,Bvrangle$ is an invariant subspace))
$endgroup$
– Enkidu
Jan 7 at 16:09
1
$begingroup$
Note that the claim is false if $dim V=0$, exclude this case. Let $U_+$, $U_-$ be the eigen spaces of $A$ and $W_+$, $W_-$ the eigen spaces of $B$. If any of the four spaces ($U_+$, say) is trivial, then then the eigen spaces of the other matrix (i.e., $W_pm$) are invariant. So all four eigen spaces have positive dimension.. $uin U_+$ can be written as $u=w_++w_-$ with $w_pmin W_pm$,the $w_pm$ depend linearly on $u$. We obtain linear maps $U_+to W_pm$, similarly $U_-to W_pm$, and ll $W$-to$U$. The kernel of any of these is nvariant, hence all 4 eigen spaces are isomorphic
$endgroup$
– Hagen von Eitzen
Jan 7 at 16:25
add a comment |
$begingroup$
I've just started learning about invariant subspaces and I came across this exercise:
Let $V$ be a finite-dimensional space over the field $mathbb{C}$. Prove that if endomorphism matrices $A$ and $B$ satisfy the equation $A^2=B^2=I$, there exists a one- or two-dimensional subspace that is both $A$ and $B$ invariant.
The only thing that I've proven thus far is that the only possible eigenvalues of $A$ and $B$ may be $1$ or $-1$ (If $v$ is an eigenvector of $A$ then $Av=lambda v Rightarrow Iv=A^2v=lambda Av=lambda^{2}v$ ; this can only happen if $lambda =pm 1$).
Any help is appreciated. Thanks.
matrices invariance
asked Jan 7 at 15:56
The CatThe Cat
18411
$endgroup$
I've just started learning about invariant subspaces and I came across this exercise:
Let $V$ be a finite-dimensional space over the field $mathbb{C}$. Prove that if endomorphism matrices $A$ and $B$ satisfy the equation $A^2=B^2=I$, there exists a one- or two-dimensional subspace that is both $A$ and $B$ invariant.
The only thing that I've proven thus far is that the only possible eigenvalues of $A$ and $B$ may be $1$ or $-1$ (If $v$ is an eigenvector of $A$ then $Av=lambda v Rightarrow Iv=A^2v=lambda Av=lambda^{2}v$ ; this can only happen if $lambda =pm 1$).
Any help is appreciated. Thanks.
matrices invariance
matrices invariance
asked Jan 7 at 15:56
The CatThe Cat
18411
asked Jan 7 at 15:56
The CatThe Cat
18411
edited Jan 7 at 16:30
The Cat
asked Jan 7 at 15:56
The CatThe Cat
18411
asked Jan 7 at 15:56
The CatThe Cat
18411
asked Jan 7 at 15:56
The CatThe Cat
18411
18411
1
$begingroup$
you need to use that an invariant subspace allows a lot more than just eigenvectors (i.e. as soon as you find an honest eigenvector $v$ for $A $ you can pimp that one into an invariant subspace (since then $langle v,Bvrangle$ is an invariant subspace))
$endgroup$
– Enkidu
Jan 7 at 16:09
1
$begingroup$
Note that the claim is false if $dim V=0$, exclude this case. Let $U_+$, $U_-$ be the eigen spaces of $A$ and $W_+$, $W_-$ the eigen spaces of $B$. If any of the four spaces ($U_+$, say) is trivial, then then the eigen spaces of the other matrix (i.e., $W_pm$) are invariant. So all four eigen spaces have positive dimension.. $uin U_+$ can be written as $u=w_++w_-$ with $w_pmin W_pm$,the $w_pm$ depend linearly on $u$. We obtain linear maps $U_+to W_pm$, similarly $U_-to W_pm$, and ll $W$-to$U$. The kernel of any of these is nvariant, hence all 4 eigen spaces are isomorphic
$endgroup$
– Hagen von Eitzen
Jan 7 at 16:25
add a comment |
1
$begingroup$
you need to use that an invariant subspace allows a lot more than just eigenvectors (i.e. as soon as you find an honest eigenvector $v$ for $A $ you can pimp that one into an invariant subspace (since then $langle v,Bvrangle$ is an invariant subspace))
$endgroup$
– Enkidu
Jan 7 at 16:09
1
$begingroup$
Note that the claim is false if $dim V=0$, exclude this case. Let $U_+$, $U_-$ be the eigen spaces of $A$ and $W_+$, $W_-$ the eigen spaces of $B$. If any of the four spaces ($U_+$, say) is trivial, then then the eigen spaces of the other matrix (i.e., $W_pm$) are invariant. So all four eigen spaces have positive dimension.. $uin U_+$ can be written as $u=w_++w_-$ with $w_pmin W_pm$,the $w_pm$ depend linearly on $u$. We obtain linear maps $U_+to W_pm$, similarly $U_-to W_pm$, and ll $W$-to$U$. The kernel of any of these is nvariant, hence all 4 eigen spaces are isomorphic
$endgroup$
– Hagen von Eitzen
Jan 7 at 16:25
1
1
$begingroup$
you need to use that an invariant subspace allows a lot more than just eigenvectors (i.e. as soon as you find an honest eigenvector $v$ for $A $ you can pimp that one into an invariant subspace (since then $langle v,Bvrangle$ is an invariant subspace))
$endgroup$
– Enkidu
Jan 7 at 16:09
$begingroup$
you need to use that an invariant subspace allows a lot more than just eigenvectors (i.e. as soon as you find an honest eigenvector $v$ for $A $ you can pimp that one into an invariant subspace (since then $langle v,Bvrangle$ is an invariant subspace))
$endgroup$
– Enkidu
Jan 7 at 16:09
1
1
$begingroup$
Note that the claim is false if $dim V=0$, exclude this case. Let $U_+$, $U_-$ be the eigen spaces of $A$ and $W_+$, $W_-$ the eigen spaces of $B$. If any of the four spaces ($U_+$, say) is trivial, then then the eigen spaces of the other matrix (i.e., $W_pm$) are invariant. So all four eigen spaces have positive dimension.. $uin U_+$ can be written as $u=w_++w_-$ with $w_pmin W_pm$,the $w_pm$ depend linearly on $u$. We obtain linear maps $U_+to W_pm$, similarly $U_-to W_pm$, and ll $W$-to$U$. The kernel of any of these is nvariant, hence all 4 eigen spaces are isomorphic
$endgroup$
– Hagen von Eitzen
Jan 7 at 16:25
$begingroup$
Note that the claim is false if $dim V=0$, exclude this case. Let $U_+$, $U_-$ be the eigen spaces of $A$ and $W_+$, $W_-$ the eigen spaces of $B$. If any of the four spaces ($U_+$, say) is trivial, then then the eigen spaces of the other matrix (i.e., $W_pm$) are invariant. So all four eigen spaces have positive dimension.. $uin U_+$ can be written as $u=w_++w_-$ with $w_pmin W_pm$,the $w_pm$ depend linearly on $u$. We obtain linear maps $U_+to W_pm$, similarly $U_-to W_pm$, and ll $W$-to$U$. The kernel of any of these is nvariant, hence all 4 eigen spaces are isomorphic
$endgroup$
– Hagen von Eitzen
Jan 7 at 16:25
add a comment |
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$begingroup$
you need to use that an invariant subspace allows a lot more than just eigenvectors (i.e. as soon as you find an honest eigenvector $v$ for $A $ you can pimp that one into an invariant subspace (since then $langle v,Bvrangle$ is an invariant subspace))
$endgroup$
– Enkidu
Jan 7 at 16:09
1
$begingroup$
Note that the claim is false if $dim V=0$, exclude this case. Let $U_+$, $U_-$ be the eigen spaces of $A$ and $W_+$, $W_-$ the eigen spaces of $B$. If any of the four spaces ($U_+$, say) is trivial, then then the eigen spaces of the other matrix (i.e., $W_pm$) are invariant. So all four eigen spaces have positive dimension.. $uin U_+$ can be written as $u=w_++w_-$ with $w_pmin W_pm$,the $w_pm$ depend linearly on $u$. We obtain linear maps $U_+to W_pm$, similarly $U_-to W_pm$, and ll $W$-to$U$. The kernel of any of these is nvariant, hence all 4 eigen spaces are isomorphic
$endgroup$
– Hagen von Eitzen
Jan 7 at 16:25