Determine for what real values of x the matrix A is positive semidefinite.

Question

Given the matrix $$A=\begin{bmatrix} 1 & (x+1) & 1 \\
(x+1) & 1 & (x+1)\\
1 & (x+1) & 1
\end{bmatrix}$$

Determine for what real values of x the matrix A is positive semidefinite.

So, I tried to find the eigenvalues of A and got the expression $$\lambda (-\lambda^2+3\lambda+(2x^2+4x))=0 \Rightarrow \lambda=0 \ \ \textrm{or} \ \ -\lambda^2+3\lambda+(2x^2+4x)=0$$

In the last equation I have $\lambda=\frac{3\pm\sqrt{8x^2+16x+9}}{2}$

Q: This is the best answer I can find as a function of $x\in\mathbb{R}$? How to write the range where $\lambda\geq0$?

Answer 1

We have

• $\det(1)=1$
• $\det\begin{bmatrix} 1 & (x+1) \\ (x+1) & 1 \end{bmatrix}=1-(x+1)^2$
• $\det A=0$

then by Sylvester criterion we need

$$1-(x+1)^2 > 0 \iff x(x+2)<0 \quad -2<x<0$$

and by direct inspection we see that $A$ is positive semidefinite also for $x=0,-2$.