For what values of $a \in \mathbb{R}$ is the the following matrix positive definite?

For what values of $a \in \mathbb{R}$ is the the following matrix positive definite?

$\begin{bmatrix} a&0&-2\\ 0&a&2\\ -2&2&a \end{bmatrix}$

Not so basically I tried to find a method to that would help me answer this question.

I started by taking $z^TIz =\begin{bmatrix} a&0&-2\\ 0&a&2\\ -2&2&a \end{bmatrix}\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}\begin{bmatrix} a&0&-2\\ 0&a&2\\ -2&2&a \end{bmatrix} = \begin{bmatrix} a^2+4&-4&-4a\\ -4&a^2+4&4a\\ -4a&4a&a^2+4 \end{bmatrix}$

Not sure what i can do from here.. I then tried looking at the determinant:

$A_1 = a$

$A_2 = a^2$

$A_3 = a^3 - 8a$

Please help, I dont know how I can find the interval nor the appropriate values of $a$.

You are already on the right track. According to Sylvester's criterion a matrix is positive definite iff all upper-left corners have positive determinant. Those are exactly the three determinants you calculated.

Let's look at them in turn:

• $A_1 = a > 0 \implies a>0$. Duh.

• $A_2 = a^2 > 0 \implies a\ne 0$. This adds nothing to the first condition.

• $A_3 = a^3-8a > 0$. Since we already know that $a>0$, we can divide that inequality by $a$ to obtain $a^2-8 > 0$ Therefore we get $a>2\sqrt{2}$.

Therefore the matrix is positive if and only if $a>2\sqrt{2}$.

The matrix is positive definite, if and only if each eigenvalue is positive.

so just find the eigenvalue. $|\lambda I-A|=0$ as $\begin{bmatrix} \lambda-a & 0 & -2\\ 0 & \lambda-a & 2\\ -2 & 2 & \lambda-a \end{bmatrix}=0$, expand it we have $(\lambda-a)^3=0$, so the eigenvalue is $\lambda_1=\lambda_2=\lambda_3=a$, we need them positive, just need $a>0$/

• This answer is not correct. You forgot to invert the signs of all 2's. The answer by @celtschk is correct as is easily verified with WA: wolframalpha.com/input/… – Bananach Mar 4 '17 at 8:08
• For $a=\sqrt{2}>0$, the vector $(1,-1,\sqrt{2})^T$ is eigenvector with eigenvalue $-\sqrt{2}<0$. – celtschk Mar 4 '17 at 8:14
• Thanks, I was wrong. the equation is $\begin{vmatrix}\lambda -a&0&2\\0&\lambda-a&-2\\2&-2&\lambda-a\end{vmatrix}=0$. then the eigenvalue is $\lambda_1 = a, \lambda_2 = a-2\sqrt{2}, \lambda_3 = a+2\sqrt{2}$, we need them all positive, so $a > 2\sqrt{2}$. thanks for point out my mistakes. – xunitc Mar 4 '17 at 13:00