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Let $b\in\mathbb{R}$ and consider the matrix

$\begin{vmatrix} 4 &-1 & b\\ -1& 2& -2\\ 0& -2& 1 \end{vmatrix}$

The problem is: for what values of $b\in\mathbb{R}$ the matrix is positive definite?

I think the correct answer is that $b=0$, because for every $b\neq 0$, the martrix is even symmetric so it makes no sense to talk about positive definiteness.

Could anyone please help me to understand if I am doing it well or I am missing something? Thank you in advance!

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  • $\begingroup$ YOu only have to check if for $b=0$ it is really positve definite. $\endgroup$ – Tito Eliatron Dec 12 '20 at 9:16
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    $\begingroup$ @TitoEliatron so my reasoning works, isn' it? The only admissible value is $b=0$ and for $b=0$ I have to check if it is positive definite or not. $\endgroup$ – C. Bishop Dec 12 '20 at 9:19
  • $\begingroup$ Your matrix cannot possibly be positive definite because it has trailing principal $2\times2$ minor is negative. $\endgroup$ – user1551 Dec 12 '20 at 9:21
  • $\begingroup$ @user1551 Sorry, question edited. $\endgroup$ – C. Bishop Dec 12 '20 at 9:22
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    $\begingroup$ On the wider reading, a non-symmetric real matrix $A$ is positive definite iff $A+A^T$ is positive definite $\endgroup$ – Henry Dec 12 '20 at 10:04
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$A$ can be decomposed in a sum $A=S+Q$ of a symmetric matrix $S=\frac 12(A+A^T)$ and a skew-symmetric matrix $Q=\frac 12(A-A^T)$.

When $A$ is symmetric, we just have $S=A$ and $Q=0$.

The general definition for definitive positiveness of a real matrix $M$ is:

  • $x^TMx>0$ for any vector $x\neq 0$.

Note: for the complex case change this to $x^*Mx$

We can notice that for skew-symmetric matrices, this product is always zero

$x^TQx\underbrace{=}_\text{this is a scalar}\left(x^TQx\right)^T=x^TQ^Tx=x^T(-Q)x=-(x^TQx)\implies x^TQx=0$

And since $\, x^TAx=x^T(S+Q)x=(x^TSx)+(x^TQx)=x^TSx\, $ then

$A$ is definite positive $\iff S$ is definite positive.

So we move on studying $\, S=\begin{bmatrix}4&-1&\frac b2\\-1&2&-2\\\frac b2&-2&1\end{bmatrix}$

We can now apply Sylvester minors criterion to the symmetric matrix $S$

$\Delta_1=\begin{vmatrix}4\end{vmatrix}=4>0$

$\Delta_2=\begin{vmatrix}4&-1\\-1&2\end{vmatrix}=8-1=7>0$

$\Delta_3=\begin{vmatrix}4&-1&\frac b2\\-1&2&-2\\\frac b2&-2&1\end{vmatrix}=-9+2b-\frac 12 b^2=-\frac 12(b^2-4b+18)<0$

Indeed the quadratic has discriminant $16-4\times 18=-56$ and leading coeff is negative.

Conclusion $A$ is never definite positive (nor semi-definite) for any value of $b$

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