MHB Is the Orthogonal Complement of an Invariant Subspace Itself Invariant?

[Sudharaka]

Hi everyone, :)

Here's a question with my answer. I would be really grateful if somebody could confirm whether my answer is correct. :)

Problem:

Prove that the orthogonal compliment \(U^\perp\) to an invariant subspace \(U\) with respect to a Hermitian transformation is itself invariant.

My Answer:

Let \(B\) denote the associated bilinear form, and \(f\) denote the Hermitian transformation. Then we have to show that, \(f(U^\perp)\subset U^\perp\). That is,

\[B(f(u'),\,u)=0\]

for all \(u\in U\) where \(u'\in U^\perp\).

Take any \(u'\in U^\perp\). Then for any \(u\in U\),

\[B(f(u'),\,u)=B(u',\,f^*(u))\]

Now since \(f\) is Hermitian (self-adjoint) we have, \(f=f^*\). Therefore,

\[B(f(u'),\,u)=B(u',\,f(u))\]

Now since \(U\) is an invariant subspace, \(f(u)\in f(U)\subset U\). Therefore,

\[B(f(u'),\,u)=B(u',\,f(u))=0\]

Am I correct? :)

 

[Opalg]

Hi everyone, :)

Here's a question with my answer. I would be really grateful if somebody could confirm whether my answer is correct. :)

Problem:

Prove that the orthogonal compliment \(U^\perp\) to an invariant subspace \(U\) with respect to a Hermitian transformation is itself invariant.

My Answer:

Let \(B\) denote the associated bilinear form, and \(f\) denote the Hermitian transformation. Then we have to show that, \(f(U^\perp)\subset U^\perp\). That is,

\[B(f(u'),\,u)=0\]

for all \(u\in U\) where \(u'\in U^\perp\).

Take any \(u'\in U^\perp\). Then for any \(u\in U\),

\[B(f(u'),\,u)=B(u',\,f^*(u))\]

Now since \(f\) is Hermitian (self-adjoint) we have, \(f=f^*\). Therefore,

\[B(f(u'),\,u)=B(u',\,f(u))\]

Now since \(U\) is an invariant subspace, \(f(u)\in f(U)\subset U\). Therefore,

\[B(f(u'),\,u)=B(u',\,f(u))=0\]

Am I correct? :)

Yes. :)

 

[Sudharaka]

Yes. :)

Thanks for the confirmation. :)