# [SOLVED]Invariant Orthogonal Compliment

#### Sudharaka

##### Well-known member
MHB Math Helper
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.

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

##### MHB Oldtimer
Staff member
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.

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

##### Well-known member
MHB Math Helper
Yes. Thanks for the confirmation. 