Complex Endomorphism $f$ with orthogonal Eigenspaces implies that $f$ is unitary
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I already posted this question earlier today without much success. After 3 hours and only 10 views I decided to give it another shot :) Of course I'll delete the second post!
Let $V ne 0$ be an finite-dimensional inner complex vectorspace and $f$ an endomorphism on said space such that there exists a $m in mathbb{N}_{ge 1}$ with $f^m=id_V$. Given that all eigenspaces are orthogonal, I have to show that $f$ is unitary. The converse I already have proven.
I thought about starting with $f^m=id_V$:
If we take an eigenvector $x$ to the eigenvalue $λ$ we get: $f(x)=λx$, therefore $f^m(x)=λ^mx$. But $f^m(x)=x$ since $f^m=id_V$. Together: $λ^mx=x$, so: $λ^m=1$.
It follows that all eigenvalues of $f$ are $m$-th roots of unity and then: $bar{lambda}=1/λ$
This is the part where I unfortunately get stuck on :P
Any Ideas?
~Cedric :)
eigenvalues-eigenvectors orthogonality roots-of-unity
asked Dec 2 '18 at 16:23
C. BrendelC. Brendel
518
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add a comment |
$begingroup$
I already posted this question earlier today without much success. After 3 hours and only 10 views I decided to give it another shot :) Of course I'll delete the second post!
Let $V ne 0$ be an finite-dimensional inner complex vectorspace and $f$ an endomorphism on said space such that there exists a $m in mathbb{N}_{ge 1}$ with $f^m=id_V$. Given that all eigenspaces are orthogonal, I have to show that $f$ is unitary. The converse I already have proven.
I thought about starting with $f^m=id_V$:
If we take an eigenvector $x$ to the eigenvalue $λ$ we get: $f(x)=λx$, therefore $f^m(x)=λ^mx$. But $f^m(x)=x$ since $f^m=id_V$. Together: $λ^mx=x$, so: $λ^m=1$.
It follows that all eigenvalues of $f$ are $m$-th roots of unity and then: $bar{lambda}=1/λ$
This is the part where I unfortunately get stuck on :P
Any Ideas?
~Cedric :)
eigenvalues-eigenvectors orthogonality roots-of-unity
asked Dec 2 '18 at 16:23
C. BrendelC. Brendel
518
$endgroup$
add a comment |
$begingroup$
I already posted this question earlier today without much success. After 3 hours and only 10 views I decided to give it another shot :) Of course I'll delete the second post!
Let $V ne 0$ be an finite-dimensional inner complex vectorspace and $f$ an endomorphism on said space such that there exists a $m in mathbb{N}_{ge 1}$ with $f^m=id_V$. Given that all eigenspaces are orthogonal, I have to show that $f$ is unitary. The converse I already have proven.
I thought about starting with $f^m=id_V$:
If we take an eigenvector $x$ to the eigenvalue $λ$ we get: $f(x)=λx$, therefore $f^m(x)=λ^mx$. But $f^m(x)=x$ since $f^m=id_V$. Together: $λ^mx=x$, so: $λ^m=1$.
It follows that all eigenvalues of $f$ are $m$-th roots of unity and then: $bar{lambda}=1/λ$
This is the part where I unfortunately get stuck on :P
Any Ideas?
~Cedric :)
eigenvalues-eigenvectors orthogonality roots-of-unity
asked Dec 2 '18 at 16:23
C. BrendelC. Brendel
518
$endgroup$
I already posted this question earlier today without much success. After 3 hours and only 10 views I decided to give it another shot :) Of course I'll delete the second post!
Let $V ne 0$ be an finite-dimensional inner complex vectorspace and $f$ an endomorphism on said space such that there exists a $m in mathbb{N}_{ge 1}$ with $f^m=id_V$. Given that all eigenspaces are orthogonal, I have to show that $f$ is unitary. The converse I already have proven.
I thought about starting with $f^m=id_V$:
If we take an eigenvector $x$ to the eigenvalue $λ$ we get: $f(x)=λx$, therefore $f^m(x)=λ^mx$. But $f^m(x)=x$ since $f^m=id_V$. Together: $λ^mx=x$, so: $λ^m=1$.
It follows that all eigenvalues of $f$ are $m$-th roots of unity and then: $bar{lambda}=1/λ$
This is the part where I unfortunately get stuck on :P
Any Ideas?
~Cedric :)
eigenvalues-eigenvectors orthogonality roots-of-unity
eigenvalues-eigenvectors orthogonality roots-of-unity
asked Dec 2 '18 at 16:23
C. BrendelC. Brendel
518
asked Dec 2 '18 at 16:23
C. BrendelC. Brendel
518
asked Dec 2 '18 at 16:23
C. BrendelC. Brendel
518
asked Dec 2 '18 at 16:23
C. BrendelC. Brendel
518
asked Dec 2 '18 at 16:23
C. BrendelC. Brendel
518
518
add a comment |
add a comment |
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