Classifying Critical Point in 3D

Question

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).

Answer 1

• 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\ $.

Answer 2

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.

2. $p, q < 0$:

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.