0

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!

zwim
  • 28,563
C. Bishop
  • 3,418
  • YOu only have to check if for $b=0$ it is really positve definite. – Tito Eliatron Dec 12 '20 at 09:16
  • 1
    @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. – C. Bishop Dec 12 '20 at 09:19
  • Your matrix cannot possibly be positive definite because it has trailing principal $2\times2$ minor is negative. – user1551 Dec 12 '20 at 09:21
  • @user1551 Sorry, question edited. – C. Bishop Dec 12 '20 at 09:22
  • On the question of whether positive definite matrices must be symmetric (or Hermitian) see https://math.stackexchange.com/questions/1954167/do-positive-semidefinite-matrices-have-to-be-symmetric - Wikipedia says "Some authors use more general definitions of definiteness, including some non-symmetric real matrices, or non-Hermitian complex ones." – Henry Dec 12 '20 at 09:44
  • 1
    On the wider reading, a non-symmetric real matrix $A$ is positive definite iff $A+A^T$ is positive definite – Henry Dec 12 '20 at 10:04
  • @Henry thank you for the comment. In my course the definition was made only for symmetric matrices. – C. Bishop Dec 12 '20 at 10:06
  • I have shown in my answer how you can go from a general matrix $A$ to the study of the symmetric part, but got no feedback, is there something you need more explanation on ? – zwim Jan 11 '21 at 22:26

1 Answers1

1

$A$ 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 $b$

zwim
  • 28,563