# Classifying Critical Point in 3D

Question: $$f(x, y, z) = px^2 +q(y^2 + z^2) +rxy + syz$$ where $$p,q,r,s \in \mathbb{R}$$ has a critical point at $$(0, 0, 0)$$. Classify this critical point. You can assume the product of $$p$$ and $$q$$ is positive. Also, $$r$$ and $$s$$ cannot be both equal to zero (either one is zero and the other is not, or neither are zero).

Attempt: I've found the Hessian matrix evaluated at the critical point $$H=\begin{bmatrix} 2p & r & 0 \\ r & 2q & s \\ 0 & s & 2q \\\end{bmatrix}$$.

I've tried to find the eigenvalues ($$\lambda$$) of $$H$$ to assess whether the point is a local minimum, maximum or saddle point, but ended up with a long messy equation that cannot be factorised easily to solve for $$\lambda$$: $$(2p - \lambda)((2q-\lambda)^2-s^2)-r^2(2q-\lambda)=0$$ which expands to $$-2r^2q+\lambda r^2 -2ps^2 +8q^2p-8\lambda qp +2\lambda ^2p+\lambda s^2 -4\lambda q^2 +4\lambda ^2 q-\lambda^3=0$$

Using Wolfram Alpha to solve this and find the eigenvalues gives these three solutions, which pushes me to consider another strategy.

So my next attempt was to see if I could classify the point via this method (see page 3) because I figured it would break it down into smaller more easier to manage equations with less unknowns, however it got pretty messy having to consider the two cases ($$p,q>0$$ and $$p,q<0$$) and then the three subcases (both nonzero $$r$$ and $$s$$, $$r=0$$ and nonzero $$s$$, nonzero $$r$$ and $$s=0$$):

• $$f_{xx}(0, 0, 0) = 2p$$
• $$\det \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}=\det \begin{bmatrix} 2p & r \\ r & 2q \end{bmatrix} = 4pq-r^2$$
• $$\det H = \det \begin{bmatrix} 2p & r & 0 \\ r & 2q & s \\ 0 & s & 2q \\\end{bmatrix} = 2p(4q^2 - s^2)-2r^2q = 8pq^2 - 2ps^2 - 2r^2 q$$

Dealing with the signs of the constants is really what's throwing me, as I am comfortable with the classification process. Any help would be greatly appreciated.

Edit: I'm pretty sure the type of critical point will be different depending on the different cases of what $$p, q, r, s$$ are (i.e. $$p,q \gt 0$$ or $$p, q \lt 0$$, and then subcases concerning $$r$$ and $$s$$ and whether they're zero or non-zero, remembering that they cannot be both zero).

• The expansion is incorrect, there should be a $\lambda^3$ somewhere. Also, it might be best to first consider the case $r = 0$, then $s = 0$, then finally $r, s \ne 0$. Commented Jun 16, 2020 at 14:57
• I believe there are errors in your characteristic polynomial for $\ H\$. I get $$(2p - \lambda)\left((2q-\lambda)^2-s^2\right)-r^2(2q-\lambda)\ .$$ If $\ p=q\$, then the eigenvalues would appear to be $\ 2q, 2q+\sqrt{r^2+s^2}\$, and $\ 2q-\sqrt{r^2+s^2}\$. Since $\ r^2+s^2>0\$, these are distinct, but $\ q\$ could be positive or negative, and $\ 2q-\sqrt{r^2+s^2}\$ could have the opposite sign to $\ q\$ when it's positive, or $\ 2q+\sqrt{r^2+s^2}\$ could have the opposite sign to it when it's negative, so it looks to me like all three types of critical point are possible. Commented Jun 16, 2020 at 16:12
• @lonzaleggiera I have corrected the characteristic equation. Commented Jun 17, 2020 at 0:27
• @mattos Correct - I have fixed this. Commented Jun 17, 2020 at 0:28

## 2 Answers

• Since $$\begin{bmatrix}x&y&z\end{bmatrix}H \begin{bmatrix}x\\y\\z\end{bmatrix}=2f(x,y,z)\ ,$$ then $$\ (0,0,0)\$$ is a local (and, in fact, global) minimum of $$\ f\$$ if and only if $$\ H\$$ is positive semi-definite. Sylvester's criterion for positive semi-definiteness tells us that a matrix is positive semi-definite if and only if all its principal minors are non-negative. The principal minors of $$\ H\$$ are the determinants of the matrices $$\begin{bmatrix}2p\end{bmatrix}, \begin{bmatrix}2q\end{bmatrix}, \begin{bmatrix}2p&r\\ r&2q\end{bmatrix}, \begin{bmatrix}2p&0\\0&2q\end{bmatrix}, \begin{bmatrix}2q&s\\ s&2q\end{bmatrix},\text{ and }H\ .$$ Since we're told that $$\ pq>0\$$, these determinants will all be non-negative if and only if $$\ p>0,q>0,\frac{r^2}{4}\le pq\$$, $$\ \frac{s^2}{4}\le q^2\$$, and $$\ 4pq^2-ps^2-r^2q\ge0\$$. These criteria can be conveniently split into three cases: \begin{align} \text{(a)}\ \ &s=0,\ p>0,\ q>0,\ \frac{r^2}{4}\le pq\\ \text{(b)}\ \ &r=0,\ p>0,\ q>0,\ \frac{s^2}{4}\le q^2\\ \text{(c)}\ \ & 0<\frac{r^2}{4}\le pq,\ 0<\frac{s^2}{4}< q^2,\ p\ge\frac{r^2q}{4q^2-s^2},\ q>0,\ \end{align}
• Likewise, $$\ (0,0,0)\$$ is a local maximum of $$\ f\$$ if and only if $$\ -H\$$ is positive semi-definite. The principal minors of $$\ -H\$$ being the determinants of the matrices $$\begin{bmatrix}-2p\end{bmatrix}, \begin{bmatrix}-2q\end{bmatrix}, \begin{bmatrix}-2p&-r\\-r&-2q\end{bmatrix}, \begin{bmatrix}-2p&0\\0&-2q\end{bmatrix}, \begin{bmatrix}-2q&-s\\ -s&-2q\end{bmatrix},\text{ and }-H\ ,$$ this will occur if and only if $$\ p<0,q<0,\frac{r^2}{4}\le pq\$$, $$\ \frac{s^2}{4}\le q^2\$$, and $$\ 4pq^2-ps^2-r^2q\le 0\$$. These criteria can again be conveniently split into three cases: \begin{align} \text{(a)}\ \ &s=0,\ p<0,\ q<0,\ \frac{r^2}{4}\le pq\\ \text{(b)}\ \ &r=0,\ p<0,\ q<0,\ \frac{s^2}{4}\le q^2\\ \text{(c)}\ \ & 0<\frac{r^2}{4}\le pq,\ 0<\frac{s^2}{4}< q^2,\ p\le\frac{r^2q}{4q^2-s^2},\ q<0,\ \end{align}
• If neither of the sets of criteria on $$\ p,q,r,s\$$ for $$\ H\$$ or $$\ -H\$$ to be positive semi-definite are satisfied, then $$\ (0,0,0)\$$ will be a saddle point of $$\ f\$$.

We have $$f(x, y, z) = \frac{1}{2} u^\mathsf{T} H u$$ where $$u = [x, y, z]^\mathsf{T}$$.

Results:

If $$p, q > 0$$, then $$(0, 0, 0)$$ is a local minimizer iff $$2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 \ge 0$$.

If $$p, q > 0$$ and $$2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 < 0$$, then $$(0, 0, 0)$$ is a saddle point.

If $$p, q < 0$$, then $$(0, 0, 0)$$ is a local maximizer iff $$-2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 \ge 0$$.

If $$p, q < 0$$ and $$-2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 < 0$$, then $$(0, 0, 0)$$ is a saddle point.

Details:

We apply the Schur complement. See https://www.cis.upenn.edu/~jean/schur-comp.pdf or https://en.wikipedia.org/wiki/Schur_complement

1. $$p, q > 0$$:

We have \begin{align} H \succeq 0 \quad &\Longleftrightarrow \quad \left( \begin{array}{cc} 2q & s \\ s & 2q \\ \end{array} \right) - \frac{1}{2p}\left( \begin{array}{c} r \\ 0 \\ \end{array} \right)\left( \begin{array}{c} r \\ 0 \\ \end{array} \right)^\mathsf{T} \succeq 0\\ \quad &\Longleftrightarrow \quad \left( \begin{array}{cc} 2q - \frac{r^2}{2p} & s \\ s & 2q \\ \end{array} \right) \succeq 0 \\ \quad &\Longleftrightarrow \quad 2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 \ge 0. \end{align} Also, $$\det H = 4pq (2q - \frac{r^2}{2p} - \frac{1}{2q}s^2)$$. If $$2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 < 0$$, then $$\det H < 0$$. Clearly, $$H$$ is not negative semidefinite since the diagonal entries are all positive. Thus, if $$2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 < 0$$, then $$H$$ has both positive and negative eigenvalues. Thus, if $$2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 < 0$$, then $$(0, 0, 0)$$ is a saddle point.

1. $$p, q < 0$$:

We have \begin{align} -H \succeq 0 \quad &\Longleftrightarrow \quad \left( \begin{array}{cc} -2q & -s \\ -s & -2q \\ \end{array} \right) - \frac{1}{-2p}\left( \begin{array}{c} -r \\ 0 \\ \end{array} \right)\left( \begin{array}{c} -r \\ 0 \\ \end{array} \right)^\mathsf{T} \succeq 0\\ \quad &\Longleftrightarrow \quad \left( \begin{array}{cc} -2q + \frac{r^2}{2p} & -s \\ -s & -2q \\ \end{array} \right) \succeq 0 \\ \quad &\Longleftrightarrow \quad -2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 \ge 0. \end{align} Also, $$\det H = 4pq (2q - \frac{r^2}{2p} - \frac{1}{2q}s^2)$$. If $$-2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 < 0$$, then $$\det H > 0$$. Clearly, $$H$$ is not positive semidefinite since the diagonal entries are all negative. Thus, if $$-2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 < 0$$, then $$H$$ has both positive and negative eigenvalues. Thus, if $$-2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 < 0$$, then $$(0, 0, 0)$$ is a saddle point.