I am struggling to prove the following (not a homework question!).
Let $A \in \mathbb{R}^{n \times n}$ be orthonormal, and $B \in \mathbb{R}^{n \times m}$. Define $C:=AB$ and $D:=A^{T}\text{Diag}(\lVert\text{row}_{1}(B)\rVert_{2},...,\lVert\text{row}_{d}(B)\rVert_{2})A$. Then $$\text{Im}(C)\subseteq\text{Im}(D)$$
I'll be honest, I am not really sure where to start. I know that the claim is equivalent to saying that for any $v \in \mathbb{R}^m$, there exists a $v^{\prime} \in \mathbb{R}^n$ such that $Cv=Dv^{\prime}$, but I failed to do this. I also see that the claim is equivalent to saying that $\text{span}(A\text{col}_{1}(B),...,A\text{col}_{m}(B))$ is a subset of the span of $D$. So it's sufficient to prove that $A\text{col}_{i}(B)$ is in the span of $D$.
Please may someone help me see how to proceed?
Thanks.
I think it's false.
Take
$$A = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}, B = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$
For $\theta = \pi/4$, we get
$$\text{im}(D) = \text{span}\left(\begin{bmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \end{bmatrix}\right) \supsetneq \text{span}\left(\begin{bmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{bmatrix}\right) = \text{im}(C)$$