# Positive definiteness of block matrices

I have a $$2 \times 2$$ block matrix of the form $$M = \left[ {\begin{array}{*{20}{c}} {\delta I}&A\\ {{A^T}}&kA \end{array}} \right]$$ where the matrix $$A$$ is positive definite and not symmetric, $$I$$ is the identity matrix, and $$k > 0$$ and $$\delta > 0$$.

1. Can we choose $$\delta > 0$$ such that the matrix $$M$$ be positive definite?

2. Is there any general formula for positive definiteness of block matrices? It seems that the Schur complement is only for symmetric matrices.

I really appreciate if anyone can help me regarding this problem.

• One could argue that positive definiteness is a property that only symmetric matrices can have. Commented Mar 4, 2023 at 19:38

## 2 Answers

There is always a $\delta$ large enought that turns $M$ positive definite.

First, since $A$ is positive definit, there is $\alpha>0$ such that $$x^TAx \ge \alpha\|x\|^2 \quad \forall x\in \mathbb R^n,$$ where I used the vector norm $\|x\|^2 = x^Tx$.

Let $x = \pmatrix{x_1\\x_2}\in \mathbb R^{2n}$. Then $$x^TMx = \delta \|x_1\|^2 + 2 x_1^T Ax_2 + k x_2^TAx_2.$$ By positive definiteness of $A$, $x_2^TAx_2 \ge \alpha \|x_2\|^2$. Now we use Cauchy-Schwarz inequality and definition of matrix 2-norm to estimate $$x_1^T Ax_2 \le \|x_1\|\cdot \|A\|\cdot \|x_2 \|.$$ Using the inequality $ab \le \frac\epsilon2 a^2 + \frac1{2\epsilon}b^2$ for all $a,b\ge 0$, we find $$2x_1^T Ax_2 \le 2\|x_1\|\cdot \|A\|\cdot \|x_2 \| \le \frac{k\alpha}2\|x_2\|^2 + \frac{2\|A\|^2}{k\alpha}\|x_1\|^2$$ Putting everything together, we find $$x^TMx = \delta \|x_1\|^2 + 2 x_1^T Ax_2 + k x_2^TAx_2 \ge (\delta - \frac{2\|A\|^2}{k\alpha}) \|x_1\|^2 + \frac{k\alpha}2\|x_2\|^2.$$ Hence, $M$ is positive if $\delta > \frac{2\|A\|^2}{k\alpha}$.

I find a solution but not sure if it's correct.

Consider

$S=[x_1 \hspace{.5cm} x_2]M \left[ \begin{array}{l} {x_1}\\ {x_2} \end{array} \right]=\delta x_1^T{x_1} + 2x_1^TA{x_2} + kx_2^TA{x_2}$

-> $S \le \delta {\left\| {{x_1}} \right\|^2} - {2\sigma _{\max }}(A)\left\| {{x_1}} \right\|\left\| {{x_2}} \right\| + k{\lambda _{\min }}(A){\left\| {{x_2}} \right\|^2}$

Hence, if we have

$\delta k{\lambda _{\min }}(A) - {4\lambda _{\max }}({A^T}A) \ge 0 \Rightarrow \delta \ge \frac{{{4\lambda _{\max }}({A^T}A)}}{{k{\lambda _{\min }}(A)}}$

then $S\ge0$.