eigenvalues of sum of matrices: $A$+block matrix are strictly less than eigenvalues of $A$+identity Let $A$ whose sum for rows is 0, can I prove that the $ \lambda_i \left ( \begin{bmatrix} I & 0\\
 0 & 0 \end{bmatrix} +A\right ) $ are strictly less than $\lambda_i \left ( \begin{bmatrix} I & 0\\
 0 & I \end{bmatrix} +A\right )$?
Any help would be greatly appreciated .
I know that $\lambda_i \left ( \begin{bmatrix} I & 0\\
 0 & I \end{bmatrix} +A\right )=1+\lambda_i(A)$ but I can't see any relation with $\lambda_i \left ( \begin{bmatrix} I & 0\\
 0 & 0 \end{bmatrix} +A\right )$.
 A: I think we can use the following inequality due to Weyl.  Let $A$ and $B$ be Hermitian matrices with eigenvalues $\lambda_i$ and $\mu_i$, respectively (indexed in descending order so that $\lambda_1$ and $\mu_1$ are the top eigenvalues).  Let $\nu_i$ denote the eigenvalues of $A+B$, indexed in the same way.  Then:
$$\nu_{i+j+1} \leq \lambda_{i+1} + \mu_{j+1}$$
for all $i$ and $j$ for which both sides of the inequality makes sense.
In your case, let $A$ be any Hermitian $2n \times 2n$ matrix and let $B = (I_n \oplus 0_n)$.  Setting $j = 0$ in Weil's inequality, we get:
$$\nu_{i+1} \leq \lambda_{i+1} + \mu_1 = \lambda_{i+1} + 1$$
The right-hand side, as you observed, is the eigenvalue of $A + I_{2n}$.  
However, you wanted the inequality to be strict.  I unfortunately don't know the full analysis of the equality case of Weyl's estimates; all I know off the top of my head is that $\nu_1 = \lambda_1 + \mu_1$ if and only if $A$ and $B$ share a principal eigenvector.  This is clearly ruled out by your condition that the row sums vanish, but I don't know what to say about the other eigenvalues.
