For what values of $a\in \mathbb{C}$, matrix $A$ is positive definite?

Question

My task is to find a value $a\in\mathbb{C}$ such that the matrix

$$A=\left(\begin{array}(1 & a & a \\ a & 1 & a\\ a & a & 1\end{array}\right)$$

is positive definite for all $\textbf{x}\in\mathbb{C}^3, \textbf{x}\neq\textbf{0}$. I tried to do the following:

$\langle A\textbf{x}, \textbf{x}\rangle = x\overline{x}+y\overline{y}+z\overline{z} + a(\overline{x}y+x\overline{y}+\overline{x}z+x\overline{z}+\overline{y}z+y\overline{z})$, where

$\textbf{x}=(x,y,z).$ Matrix $A$ is p.d. if $\langle A\textbf{x}, \textbf{x}\rangle>0$ for all non-zero vectors. I get:

$\langle A\textbf{x}, \textbf{x}\rangle = \langle\textbf{x},\textbf{x}\rangle + a(\overline{x}y+x\overline{y}+\overline{x}z+x\overline{z}+\overline{y}z+y\overline{z})$. I know $\langle\textbf{x},\textbf{x}\rangle\geq 0$, so now I must find $a$ such that

$a(\overline{x}y+x\overline{y}+\overline{x}z+x\overline{z}+\overline{y}z+y\overline{z})>0$. The sum inside the brackets is a real number $r=\overline{x}y+x\overline{y}+\overline{x}z+x\overline{z}+\overline{y}z+y\overline{z}$. So I need $a$ such that

$$ar>0,$$

By selecting $a=r$, I get $r^2>0$ but now $a$ is dependent on $r$ and I'm stuck. How to proceed?

Answer 1

We have: $$ A = a\cdot\begin{pmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix}+(1-a)\cdot I $$ where the matrix made by $1s$ only is a rank-$1$ matrix with eigenvalues $3,0,0$.

It follows that the eigenvalues of $A$ are $2a+1,1-a,1-a$, so $A$ is positive definite iff $2a+1>0$ and $1-a>0$, hence iff $\color{red}{-\frac{1}{2}<a<1}$.

Answer 2

The determinant is $\det A=1+2a^3-3a^2$ and by Sylvester's criterion the conditions are $$ \begin{cases} 1-a^2>0\\[4px] 2a^3-3a^2+1>0 \end{cases} $$ as the first principal minor is positive.

Since the second inequality is the same as $(a-1)^2(2a+1)>0$ and the solutions for the first one are $-1<a<1$, we get the final condition $\displaystyle-\frac{1}{2}<a<1$