Distorted Unitary matrices

Question

Let $U$ be an unitary and $D$ be a diagonal matrix. We know that for all vectors $v$ on the sphere $Uv$ is on the sphere and, $$\langle Uv,Uv\rangle=\langle v,v\rangle.$$ What are the vectors $v$ on the sphere at origin such that $$\langle DUv,DUv\rangle=\langle v,v\rangle?$$

Answer 1

Since $U$ is unitary, $UU^t=Id$, hence your question is equivalent to ask for vectors s.t. $\langle Dv,Dv\rangle=\langle v,v\rangle$. Since $D$ is diagonal, this is the case for vectors that are eigen-vectors of $D$ with eigenvalue $\pm1$.

Answer 2

Note that $$ \langle DUv,DUv\rangle = (DUv)^*DUv = (Uv)^*D^*D(Uv)^* = (Uv)^*|D|^2(Uv)^* $$ Setting $w = Uv$, this is just $w^*|D|^2w$. So, noting that $\langle w,w \rangle = \langle v,v \rangle$ the question can be rephrased as "when does $w^*|D|^2w = \langle w,w \rangle$"?

Now, suppose $w$ is given by $w = \pmatrix{w_1 & \cdots & w_n}^T$ and $D$ is given by $$ D = \pmatrix{d_{11}&&\\&\ddots&\\&&d_{nn}} $$ Then we have $$ w^*|D|^2w - w^*w = w^*(|D|^2-I)w= (|d_{11}|^2-1)|w_1|^2 + \cdots + (|d_{nn}|^2-1)|w_n|^2 $$ The above will be equal to zero if $w_i = 0$ for all $i$ such that $|d_{ii}|\neq 1$.