Find the sign of eigenvalues generated from Hessian matrix

Question

I have to classify a critical point of a function of 3 variables. I determined the Hessian and I know I can orthogonally diagonalise this, so to classify the critical point, I only need to know the signs of the eigenvalues. However, in my function I do not know how to go any further with the problem. I have the following values for my eigenvalues at the moment: $$\lambda_1 = 2b, \lambda_{2,3}= b + d \pm \sqrt{(a-b)^2 +c^2 +d^2}$$ I know that $a \times b> 0$ and that $c$ and $d$ cannot simultaneously equal to $0$ and since the eigenvalues must be real, I also know that $$(a-b)^2 +c^2 +d^2 > 0$$ How can I further this problem and work out the signs of the eigenvalues?

Edit: The original function is $$f(s,t,u) = as^2 + b(t^2 + u^2) +cst + dsu$$ and the Hessian from this function is: $$\left[ \begin{matrix} 2a & c & d \\ c & 2b & 0 \\ d & 0 & 2b \\ \end{matrix}\right]$$

Answer 1

I assume WLOG that $b\gt 0$ (if not you should be able to consider the negative of your matrix -- the below analysis then follows-- then negate at the end)

For suitable permutation matrix $P$ we have
$A:=\left[ \begin{matrix} 2b & 0 & d \\ 0 & 2b & c \\ d & c & 2b \\ \end{matrix}\right]=P^T\left[ \begin{matrix} 2a & c & d \\ c & 2b & 0 \\ d & 0 & 2b \\ \end{matrix}\right]P$
$A$ is congruent (and similar) to your matrix so it has the same signature.

Applying e.g. Sylvester's determinant criterion to $A$, we see that the leading $1\times 1$ and $2\times 2$ principal submatrices have positive determinant hence
$\mathbf 0 \prec A$ iff $0\lt \det\big(A\big)=8ab^2-2bc^2-2bd^2=2b\Big(4ab -\big(c^2+d^2\big)\Big)$ i.e. iff
$4ab \gt c^2+d^2$
depending on how well you understand the proof of Sylvester's Determinant Criterion, or alternatively via application of Cauchy Eigenvalue interlacing, you get signatures of:

i.) $\big(3,0\big)$ if $4ab \gt c^2+d^2$
ii.) $\big(2,0\big)$ if $4ab = c^2+d^2$
iii.) $\big(2,1\big)$ if $4ab \lt c^2+d^2$

Answer 2

Sylvester's Determinant Criterion is a very effective method, as an alternative with $\alpha = \frac a b$, $\gamma=\frac c b$ and $\delta=\frac d b$ we have

$$\frac1{2b}x^TAx= \alpha x^2+y^2+z^2+\gamma xy+\delta xz=\left(\alpha-\frac{\gamma^2}4-\frac{\delta^2}4\right)x^2+\left(y+\frac \gamma 2x\right)^2+\left(z+\frac \delta 2x\right)^2$$

therefore for $b>0$ the signature is

• $(n_+,n_-,n_0)=(3,0,0)$ for $\alpha-\frac{\gamma^2}4-\frac{\delta^2}4 >0 \implies 4ab>c^2+d^2$
• $(n_+,n_-,n_0)=(2,1,0)$ for $\alpha-\frac{\gamma^2}4-\frac{\delta^2}4 <0 \implies 4ab<c^2+d^2$
• $(n_+,n_-,n_0)=(2,0,1)$ for $\alpha-\frac{\gamma^2}4-\frac{\delta^2}4 =0 \implies 4ab=c^2+d^2$

and for $b<0$ the signature is

• $(n_+,n_-,n_0)=(0,3,0)$ for $\alpha-\frac{\gamma^2}4-\frac{\delta^2}4 >0 \implies 4ab>c^2+d^2$
• $(n_+,n_-,n_0)=(1,2,0)$ for $\alpha-\frac{\gamma^2}4-\frac{\delta^2}4 <0 \implies 4ab<c^2+d^2$
• $(n_+,n_-,n_0)=(0,2,1)$ for $\alpha-\frac{\gamma^2}4-\frac{\delta^2}4 =0 \implies 4ab=c^2+d^2$