Is the Orthogonal Complement of an Invariant Subspace Itself Invariant?

In summary, the conversation involved a question and answer about proving the invariance of an orthogonal complement with respect to a Hermitian transformation. The answer used the associated bilinear form and self-adjoint property of the transformation to show that the orthogonal complement is indeed invariant. The person asking for confirmation received it and thanked for it.
  • #1
Sudharaka
Gold Member
MHB
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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? :)
 
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  • #2
Sudharaka said:
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. :)
 
  • #3
Opalg said:
Yes. :)

Thanks for the confirmation. :)
 

FAQ: Is the Orthogonal Complement of an Invariant Subspace Itself Invariant?

What is an Invariant Orthogonal Compliment?

An Invariant Orthogonal Compliment (IOC) is a vector space that is perpendicular to a given vector space and remains unchanged under linear transformations.

How is an Invariant Orthogonal Compliment different from an orthogonal complement?

An IOC is a type of orthogonal complement, but it is unique because it remains invariant under linear transformations whereas a general orthogonal complement may not.

What is the importance of Invariant Orthogonal Compliments in linear algebra?

IOCs allow for the decomposition of a vector space into two complementary subspaces, making it easier to analyze and understand linear transformations.

How can Invariant Orthogonal Compliments be calculated?

IOCs can be calculated using the Gram-Schmidt process, which involves finding an orthogonal basis for the given vector space and then extending it to form a basis for the IOC.

Are Invariant Orthogonal Compliments unique?

Yes, IOCs are unique for a given vector space and linear transformation. However, there may be multiple different IOCs for different linear transformations on the same vector space.

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