A question about comparison between orthogonal projections

Question

$P\in\mathbb{R}^{n\times n}$ is an orthogonal projection, if $P^2=P$ and $P^T=P$. Let $X\in\mathbb{R}^{n\times d}$ such that $\text{rank}(X)=n$. Now, assume that $P_1$ and $P_2$ are orthogonal projections such that $$\text{Im}(P_1)\subseteq\text{span}(X)\qquad\text{and}\qquad\text{rank}(P_1)=m,$$ $$\text{Im}(P_2)\subseteq\text{span}(X)\qquad\text{and}\qquad\text{rank}(P_2)=r,$$ where $m\leq r\leq n$. Which of the following relations is true? $$\Vert P_1X\Vert_F\leq \Vert P_2X\Vert_F\qquad\text{or}\qquad \Vert P_2X\Vert_F\leq \Vert P_1X\Vert_F,$$ where $\Vert\cdot\Vert_F$ is the Frobenius norm.

Answer 1

Since $\|P_1X\|_F$ and $\|P_2X\|_F$ are numbers, you will always have one of the two inequalities. With the conditions you specified, each could occur, by taking $m=r$ and switching roles.

To make this more concrete, let $n=d=3$, $m=1$, $r=2$. Let $$ X=\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix},\ \ P_1=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix},\ \ P_2=\begin{bmatrix}0&0&0\\0&1&0\\0&0&1\end{bmatrix}. $$ Then $$ \|P_1X\|_F=1,\ \ \|P_2X\|_F=\sqrt{13}. $$ If now $$ X=\begin{bmatrix}1&0&0\\0&1&0\\0&0&4\end{bmatrix},\ \ P_1=\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix},\ \ P_2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}, $$ then $$ \|P_1X\|_F=4,\ \ \|P_2X\|_F=\sqrt2. $$