Let $A \in M_{n,p}(\mathbb{R})$ such that the rank of $A$ is $k \leq \min(n,p)$. Moreover let $M \in M_{n,p}(\mathbb{R})$ be a matrix of rank $l$ with $l+1 \leq k$. We denote $(\lambda_i)_{i \leq k}$ the square of the singular values of $A$ and make the extra assumption that : $\lambda_l > \lambda_{l+1}$.
Prove that :
$$ \| M - A \|_F^2 \geq \sum_{h = l+1}^k \lambda_h$$ and find for which matrix $M$ the equality occurs.
N.B : $\| \cdot \|_F$ is the Frobenius norm (ie. $\| AB \|_F = tr(^tAB)$).
Let's be honest : I don't know at all where to begin. There are a lot of assumptions on the matrix in this problem and I don't know what is going on. I don't have intuition of what I need to prove, and why we are making these assumptions on the matrix.
Since the assumptions are on the square of the singular values of $A$ I guess that a good start would be to use the singular value decomposition of $A$and then maybe somehow there is a tricky way using the trace to get the square of the singular values...
Thank you !