# For what values of $b\in\mathbb{R}$ this matrix is positive definite?

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.

• YOu only have to check if for $b=0$ it is really positve definite. – Tito Eliatron Dec 12 '20 at 9:16
• @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 9:19
• Your matrix cannot possibly be positive definite because it has trailing principal $2\times2$ minor is negative. – user1551 Dec 12 '20 at 9:21
• @user1551 Sorry, question edited. – C. Bishop Dec 12 '20 at 9:22
• 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

$$A$$ 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 $$b$$