# Inequality about Eigenvalues of Symmetric Block Matrix

Let $M = \left( \begin{array}{cc} A & B \\ B^T & C \end{array} \right)$ be a real symmetric matrix ($A, C$ are symmetric matrices). Let $\lambda_m$ and $\lambda_M$ denote the minimum and maximum eigenvalues of $M$, and $\mu_A$ and $\mu_C$ denote, respectively, the maximum eigenvalues of $A$ and $C$.

Show that $\lambda_m+\lambda_M \leq \mu_A + \mu_C$.

I've just got a hint from others: consider the positive semidefinite matrix $M - \lambda_m I$. Maybe there are other methods.

• Are you familiar with the Schur complement formula? Nov 23, 2017 at 16:51

Here I follow the hint and provide an answer. Let's first assume that $M$ is positive semi-definite and show that $\lambda_M \leq \lambda_A + \lambda_B$.

Let $z=\left[\begin{array}{c}x\\y\end{array}\right]\in \mathbb{R}^n$, $x\in \mathbb{R}^s, y\in \mathbb{R}^t$, $s, t$ are the order of matrices $A$ and $C$.

Take a unit eigenvector $z_M=(x_M, y_M)$ of $M$ associated with $\lambda_M$. Then we have $$\lambda_M=z_M^T Mz_M=x_M^T A x_M + 2x_M^T B y_M + y_M^T Cy_M.$$ Note that $A$ and $C$ are now also positive semi-definite, implying that $\lambda_A\geq 0$ and $\lambda_C\geq 0$.

If $x_M=0$, then we have $y_M^Ty_M=1$ and $\lambda_M \leq y_M^T CY_M \leq \lambda_C y_M^Ty_M = \lambda_C \leq \lambda_A+\lambda_C$.

If $y_M=0$, then we have $x_M^T x_M=1$ and $\lambda_M \leq x_M^T A x_M \leq \lambda_A x_M^Tx_M = \lambda_A \leq \lambda_A+\lambda_C$.

Now we assume that $x_M\neq 0, y_M\neq 0$. Consider $z(s,t)=\left[\begin{array}{c}sx_M\\ty_M\end{array}\right]\in \mathbb{R}^n$.

By the assumption of positive semi-definteness, we have $$0\leq s^2 x_M^T A x_M +2st x_M^TBy_M+t^2 y_M^T C y_M.$$ Combining these two yields $$\lambda_M \leq (1+s^2) x_M^T A x_M +2(st+1) x_M^TBy_M+ (1+t^2) y_M^T C y_M.$$ Take $s=\sqrt{\dfrac{y_M^Ty_M}{x_M^T x_M}}$ and $t=-\sqrt{\dfrac{x_M^T x_M}{y_M^Ty_M}}$. Thus, $st=-1$, eliminating the cross terms, $$\lambda_M \leq \frac{x_M^T A x_M}{x_M^T x_M} + \frac{y_M^T C y_M}{y_M^T y_M}\leq \lambda_A+\lambda_C.$$ In general, $M$ may not be positive semi-definite, and we consider $M'=M-\lambda_m I$. Here we have $$\lambda_M'=\lambda_M-\lambda_m,\quad\lambda_A'=\lambda_A-\lambda_m, \quad \lambda_C'=\lambda_C-\lambda_m.$$ Applying the above proven proposition to $M'$, we have $\lambda_M'\leq \lambda_A'+\lambda_C'$, that is $$\lambda_M-\lambda_m\leq (\lambda_A-\lambda_m)+(\lambda_C-\lambda_m)\Rightarrow \lambda_M +\lambda_m \leq \lambda_A +\lambda_C.$$