LS solution with orthonormal matrix

Question

Let $A\in{ \mathbb{R}_{m \times n} }$ matrix, and orthonormal matrix $ P\in{ \mathbb{R}_{m \times m}}$. Let $ \underline{b} \in \mathbb{R}_{m} $.

Can I say the following?

1. $\min||A\underline{x}-\underline{b}||_2^2 $ has single solution $\iff$ $\min||PA\underline{x}-\underline{b}||_2^2 $ has single solution?

2. $\min||A\underline{x}-\underline{b}||_2^2 = 0$ $\iff$ $\min||PA\underline{x}-\underline{b}||_2^2 = 0 $?

For both questions I got that $\min||PA\underline{x}-\underline{b}||_2^2 = \min ||A\underline{x} - P^T\underline{b}||_2^2$ but it doesn't seems to help very much... I also tried to simplify the LS problem by 2-norm definition by writing $ (PA\underline{x}-\underline{b})^T(PA\underline{x}-\underline{b})$ but I can't see how it helps.

Answer 1

1. Yes. A LS problem $\min_x\|Ax-b\|_2$ has a unique solution if and only if $A$ has full column rank. Premultiplying $A$ with an orthogonal matrix does not change its rank.

2. No. For example, $$ \min_x\left\|\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}x-\begin{bmatrix}0\\1\\0\end{bmatrix}\right\|_2=0, $$ but $$ \min_x\left\|\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}x-\begin{bmatrix}0\\1\\0\end{bmatrix}\right\|_2 = \min_x\left\|\begin{bmatrix}1&0\\0&0\\0&1\end{bmatrix}x-\begin{bmatrix}0\\1\\0\end{bmatrix}\right\|_2=1\neq 0, $$