- #1
Sudharaka
Gold Member
MHB
- 1,568
- 1
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? :)
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? :)