Suppose T ∈ L(V) and U is a subspace of V. Prove that U is(adsbygoogle = window.adsbygoogle || []).push({});

invariant under T if and only if U⊥ is invariant under T∗.

Now for reference, L(V) is the set of transformations that map v (a vector) from V to V.

T* is the adjoint operator.

The case where the dimension of U is less than V bugs me. How can U^{⊥}

be invariant under T* when T* maps from U^{⊥}to V unless mapping to V

also counts as mapping to U^{⊥}since U^{⊥}is in V itself. Now when

I say map to V, I mean lets say V is 3 dimensional and U^{⊥}is 1 dimensional.

Then u=(x3) gets mapped (by T*) to a vector with three nonzero components.

Does this count as mapping from U^{⊥}to U^{⊥}?

I'm just a bit confused about this, and any help will be greatly appreciated.

I guess that I'm confused enough that my post doesn't make much sense,

so in that case, could someone nudge me in the right direction (ie give a good hint)?

Thank you!

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# Question about a proof.

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