Finding all $\alpha$ such that a matrix is positive definite

Question

I have

$A = $ $ \left[\begin{array}{rrr} 2 & \alpha & -1 \\ \alpha & 2 & 1 \\ -1 & 1 & 4 \end{array}\right] $

and I want to find all $\alpha$ such that $A$ is positive definite.

I tried

$ x^tAx = $ $ \left[\begin{array}{r} x & y & z \end{array}\right] $ $ \left[\begin{array}{rrr} 2 & \alpha & -1 \\ \alpha & 2 & 1 \\ -1 & 1 & 4 \end{array}\right] $ $ \left[\begin{array}{r} x \\ y \\ z \end{array}\right] $

$=$ $ \left[\begin{array}{r} 2x + \alpha y - z & \alpha x + 2y + z & -x + y + 4z \end{array}\right] $ $ \left[\begin{array}{r} x \\ y \\ z \end{array}\right] $

$= 2x^2 + \alpha xy - xz + \alpha xy + 2y^2 + yz - xz + yz + 4z^2$

$= 2 \alpha xy + 2x^2 + 2y^2 - 2xz + 2yz + 4z^2$

and I wanted to solve the inequality $2 \alpha xy + 2x^2 + 2y^2 - 2xz + 2yz + 4z^2 > 0$ for $\alpha$, but I wasn't sure what to do next.

Am I doing this correctly?

Answer 1

Maybe this helps:

A hermitian matrix is positive definite $\Leftrightarrow$ all leading principal minors are positive.

So $\left|\begin{array}{r} 2 \end{array}\right|$, $\left|\begin{array}{rr} 2 & \alpha \\ \alpha & 2 \end{array}\right|$ and $\left|\begin{array}{rrr} 2 & \alpha & -1 \\ \alpha & 2 & 1 \\ -1 & 1 & 4 \end{array}\right|$ have to be positive.

This gives us $$\begin{align}&\text{I:}\quad 2>0\\ &\text{II:}\quad 4-\alpha^2>0\\ &\text{III:}\quad -4\alpha^2 - 2\alpha +12>0 \end{align} $$ Try to solve this!

Answer 2

The matrix is positive definite if and only if their eigenvalues are positive, so we calculate the characteristic polynomial $\chi_A(x)=\det(xI-A)$ and we solve for $x$ and we find: $$\lambda_1=\frac{1}{2}\sqrt{\alpha^2+4\alpha+12}-\frac{1}{2}\alpha+3,\quad\lambda_2=3-\frac{1}{2}\sqrt{\alpha^2+4\alpha+12}- \frac{1}{2}\alpha,\quad \lambda_3=\alpha+2$$ hence we should find the $\alpha$ for which $\lambda_i>0, i=1,2,3$.