Relations between the second norms of matrices Consider two arbitrary complex matrices $L$ and $M$. Let $\|L\|_2 \le \|M\|_2$. Prove that
$$
        \Bigg\|\begin{bmatrix}
        I & 0 \\
        L & I \\
        \end{bmatrix}\Bigg\|_2 \le
        \Bigg\|\begin{bmatrix}
        I & 0 \\
        M & I \\
        \end{bmatrix}\Bigg\|_2
$$
My attempt was to reduce the problem. Namely, let $L = U\Sigma V^*$ be the SVD for $L$. Then it is easy to see that
$$
\begin{bmatrix}
I & 0 \\
L & I \\
\end{bmatrix}
\begin{bmatrix}
I & L^* \\
0 & I \\
\end{bmatrix} = 
\begin{bmatrix}
V & 0 \\
0 & U \\
\end{bmatrix}
\begin{bmatrix}
I & \Sigma \\
\Sigma & \Sigma^2 + I \\
\end{bmatrix}
\begin{bmatrix}
V^* & 0 \\
0 & U^* \\
\end{bmatrix}
$$
It means that the eigenvalues of $ \begin{bmatrix}
I & 0 \\
L & I \\
\end{bmatrix}
\begin{bmatrix}
I & L^* \\
0 & I \\
\end{bmatrix} $ and $ \begin{bmatrix}
I & 0 \\
\Sigma & I \\
\end{bmatrix}
\begin{bmatrix}
I & \Sigma \\
0 & I \\
\end{bmatrix} $ are equal, so the singlular values for $ \begin{bmatrix}
I & 0 \\
L & I \\
\end{bmatrix} $ and $ \begin{bmatrix}
I & 0 \\
\Sigma & I \\
\end{bmatrix} $.
It follows that it is sufficient to prove the statement for diagonal matrices with nonnegative values on the diagonal.
But here I got stuck. Thanks for any help or ideas!

I will handle only the square matrix case here.  The other cases are similar, but require some care.
Let's reframe the question: if $L$ and $M$ are such that the largest singular values satisfy $\sigma_1(L) \geq \sigma_1(M)$, then (it suffices to show that) the diagonal matrices $\Sigma_M,\Sigma_N$ of singular values are such that the largest eigenvalue of 
$
\pmatrix{I & \Sigma_L \\
\Sigma_L & \Sigma_L^2 + I}
$
is greater than the largest eigenvalue of 
$
\pmatrix{I & \Sigma_M \\
\Sigma_M & \Sigma_M^2 + I}
$
To that end, suppose $\Sigma$ is a matrix with diagonal entries $\sigma_1 \geq \cdots \geq \sigma_n$.  Let 
$$
A = \pmatrix{I & \Sigma \\
\Sigma & \Sigma^2 + I} = I + 
\pmatrix{0 & \Sigma \\
\Sigma & \Sigma^2}
$$
Now, apply a permutation matrix similarity.  We reach the block-diagonal matrix given by
$$
PAP^* = I + 
\pmatrix{B_1 \\ & B_2 \\ && \ddots \\ &&& B_n}
$$
where $B_i =\left[\begin{smallmatrix}0&\sigma_i\\ \sigma_i & \sigma_i^2\end{smallmatrix}\right]$.
Each block of this has the characteristic equation $\lambda^2 - \sigma_i^2\lambda - \sigma_i^2$.  It follows that the eigenvalues of $B_i$ are 
$$
\lambda = \frac{\sigma_i^2 \pm \sqrt{\sigma_i^4 + 4\sigma_i^2}}{2} = 
\frac{\sigma_i^2 \pm \sigma_i\sqrt{\sigma_i^2 + 4}}{2}
$$
So, the largest eigenvalue of $A$ is $1 + \frac 12\left(\sigma_1^2 + \sigma_1\sqrt{\sigma_1^2 + 4}\right)$.  Now, it suffices to note that if $\sigma_1(L) \geq \sigma_1(M)$, then
$$
1 + \frac 12\left(\sigma_1^2(L) + \sigma_1(L)\sqrt{\sigma_1^2(L) + 4}\right) \geq
1 + \frac 12\left(\sigma_1^2(M) + \sigma_1(M)\sqrt{\sigma_1^2(M) + 4}\right)
$$
