Orthogonal Linear Transformation of Subspace

Question

given that U is an orthogonal n by n matrix, and linear transformation T map from $R^n$ to $R^n$ be $T(x) = Ux$. Let W be a subspace of $R^n$ such that $T(W) \subseteq W$. Prove $T(W) = W$ and $T(W^\perp) = W^\perp$. Here is my attempt:

For the first question, I constructed an orthogonal basis U for W, and showed that $T(U)$ is orthogonal and linearly independent with dimension dim(W).

Any hint will be appreciated. Thanks in advance.

Answer 1

If $w\in T(W^\bot)$ then there is some $v\in W^\bot$ such that $T(v)=Uv=w$. Take any $u\in W$.

Orthogonal matrices preserve inner product, just note that $\langle a,b\rangle=b^\mathsf t a=b^\mathsf tU^{-1}Ua=b^{\mathsf t}U^{\mathsf t}Ua=(Ub)^{\mathsf t}(Ua)=\langle Ua,Ub\rangle$,

and $U^{-1}u\in W$, since

$T^{-1}(T(W))=W$ because of $T$ being bijective ($U$ is invertible),

so we have that

$0=\langle U^{-1}u,v\rangle=\langle U(U^{-1}u),Uv\rangle=\langle u,w\rangle$,

so $w\in W^\bot$, and therefore $T(W^\bot)\subseteq W^\bot$.

You can do something similar for the other inclusion, and then you get the equality.

Answer 2

Just curious- For the first part is it sufficient to prove T is injective to say T(W)=W. Since T being injective implies kernel is a nullity, which implies T is surjective and hence a bijection?

Thanks!