Let $V$ be a vector space and $V_1 \subset V_2$ two subspaces of $V$. Then, denoting by $P_1$ and $P_2$ the orthogonal projectors on $V_1$ and $V_2$ respectively, it holds for any matrix $A$ of appropriate dimension $$ \|P_1A\|_F \leq \|P_2 A\|_F, $$ and $$ \|(I - P_1)A\|_F \geq \|(I-P_2) A\|_F, $$ where $\|\cdot\|_F$ is the Frobenius matrix norm.
Any idea on how to prove this result?
EDIT (following the comment of Bertrand)
For a vector $v$ it is well known that $$\|P_1 v\|_2 \leq \|P_2 v\|_2.$$ From the definition of the Frobenius norm we have $$\|P_1 A \|_F^2 = \sum_{i,j}a_{ij}^2 = \sum_{i} \|(P_1A)_j\|_2^2 = \sum_{i} \|P_1A_j\|_2^2,$$ where $(PA)_j$ and $A_j$ are the columns of $PA$ and $A$ respectively. Together with the inequality above, this implies $$\|P_1 A \|_F^2 \leq \sum_i \|P_2A_j\|_2^2 = \|P_2 A\|_F^2,$$ which is the desired result.