Is the Orthogonal Complement of a U-Invariant Subspace Also U-Invariant?

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Homework Statement

Let U be a unitary operator on an inner product space V, and let W be a finite-dimensional U-invariant subspace of V. Prove that

(a) U(W) = W
(b) the orthogonal complement of W is U-invariant
(for ease of writing let the orthogonal complement of W be represented by W^.

Homework Equations

Unitary: U*U = UU* = I

The Attempt at a Solution

(a) first show that U(W) is contained in W, and then show that W in contained in U(W).
- U(W) is contained in W because W is U-invariant
- show W is contained in U(W)
choose x in W and show it is contained in U(W)
U(x) is in W
Is this circular thinking?
Should I instead show that the range of U (restricted to W) is W itself? Or that the nullity of U (restricted to W) is 0?

(b)
Note: U restricted to W (let’s call it U_w) is also unitary.

W^ = {x in V : <x, y>=0 for all y in W}

Now show that U(W^) is contained in W^

I’m not sure what to do now.

Thanks for your help!

 

[AKG]

Homework Statement

Let U be a unitary operator on an inner product space V, and let W be a finite-dimensional U-invariant subspace of V. Prove that

(a) U(W) = W
(b) the orthogonal complement of W is U-invariant
(for ease of writing let the orthogonal complement of W be represented by W^.

Homework Equations

Unitary: U*U = UU* = I

The Attempt at a Solution

(a) first show that U(W) is contained in W, and then show that W in contained in U(W).
- U(W) is contained in W because W is U-invariant
- show W is contained in U(W)
choose x in W and show it is contained in U(W)
U(x) is in W
Is this circular thinking? Should I instead show that the range of U (restricted to W) is W itself? Or that the nullity of U (restricted to W) is 0?

Do the last one.

(b)
Note: U restricted to W (let’s call it U_w) is also unitary.

W^ = {x in V : <x, y>=0 for all y in W}

Now show that U(W^) is contained in W^

I’m not sure what to do now.

Thanks for your help!

Use the definitions. You want to say that if x is in W^, then so is U(x). Well x is in W^ iff for all y in W, <x,y> = 0, and U(x) is in W^ iff for all y' in W, <U(x),y'> = 0.