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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?

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    $\begingroup$ I'd use determinants here instead of quadratic forms like this. Specifically, look at the its leading principal minors: the determinants of 2 (trivial), $\begin{bmatrix}2&\alpha\\\alpha&2\end{bmatrix}$, and the full matrix. All three must be positive. $\endgroup$ – Michael Grant Apr 7 '13 at 13:13
  • $\begingroup$ Do you want an example of $\alpha$ for which the matrix is positive definite or all the possible $\alpha$? $\endgroup$ – user63181 Apr 7 '13 at 13:14
  • $\begingroup$ @SamiBenRomdhane I'm looking for an interval where $A$ is positive definite. $\endgroup$ – badjr Apr 7 '13 at 13:15
  • $\begingroup$ user38034's hint will lead you to what you want, if you know how to determine where a quadratic function is positive. $\endgroup$ – Julien Apr 7 '13 at 13:19
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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!

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    $\begingroup$ Great hint, +1. Minor details: a "hermitian" matrix is positive definite if and only if all of the "leading" principal minors are positive (you don't need all the principal minors, like you don't need $4$, for instance, as you perfectly did in your answer). This is called Sylvester's criterion. $\endgroup$ – Julien Apr 7 '13 at 13:18
  • $\begingroup$ Thanks, edited the word "leading" $\endgroup$ – Jakube Apr 7 '13 at 13:21
  • $\begingroup$ You're welcome. You should add hermitian too. It is false otherwise. $\endgroup$ – Julien Apr 7 '13 at 13:22
  • $\begingroup$ How did you get inequality III? $\endgroup$ – user13985 Aug 4 '18 at 21:32
  • $\begingroup$ @user13985 Determinant of the third matrix > 0. $\endgroup$ – Jakube Aug 5 '18 at 12:13
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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 $\lambda_i>0, i=1,2,3$.

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  • $\begingroup$ I suspect that if you eliminate the square root in either $\lambda_1>0$ or $\lambda_2>0$ you will obtain a quadratic form similar to the determinant above. $\endgroup$ – Michael Grant Apr 7 '13 at 14:10

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