Is the Hessian symetric for $z^3+y^2+xy+yz+3x-3z$?

Question

I wish to study the function $f(x,y,z) = z^3+y^2+xy+yz+3x-3z$ and find its extreme values.

I search for values for which $\nabla f(x,y,z) =0$.

$$\frac{\partial f}{\partial x} = y+3$$ $$\frac{\partial f}{\partial y} = 2y+x+z$$ $$\frac{\partial f}{\partial z} = 3z^2+y-3$$

I wanted to see if the Hessian is positive definite, but when I compute it I get:

$$H=\begin{pmatrix} \frac{\partial^2f}{\partial x\partial x}& \frac{\partial^2f}{\partial x\partial y} & \frac{\partial^2f}{\partial x\partial z}\\ \frac{\partial^2f}{\partial y\partial x}& \frac{\partial^2f}{\partial y\partial y} & \frac{\partial^2f}{\partial y\partial z} \\ \frac{\partial^2f}{\partial z\partial x}& \frac{\partial^2f}{\partial z\partial y} & \frac{\partial^2f}{\partial z\partial z}\end{pmatrix}$$

$$H=\begin{pmatrix}0& 1 & 0\\ 1 & 2 & 1 \\ 0& 1 & 6z\end{pmatrix}$$

However, it is not symetric.

• Did I compute the Hessian correctly?
• If it is in fact not symetric, what can I do to determine if the values are extreme values?

Answer 1

I hope you already have calculated the two critical points. It is the solution of your three equations. Your (symmetric) Hessian Matrix is right.

You have to calculate the eigenvalues of $\textbf H$. The equation is $det(\textbf H-\lambda \textbf I)=0$

$\textbf I$ is the identity matrix and $\lambda$ is a unknwon scalar. In this case the values of the eigenvalues.

Thus the equation is $-\lambda\cdot (2-\lambda)\cdot (6\cdot z-\lambda)-6\cdot z=0$

The critical values for $z$ are $\sqrt 2$ and $-\sqrt 2$. You get two equations

$-\lambda\cdot (2-\lambda)\cdot (6\cdot \sqrt 2 -\lambda)-6\cdot\sqrt 2=0 \qquad (1)$

$-\lambda\cdot (2-\lambda)\cdot (-6\cdot \sqrt 2 -\lambda)+6\cdot \sqrt 2=0 \qquad (2)$

Calculate for each equation the (three) solutions.

If $\lambda_1,\lambda_2$ and $\lambda_3$ are all positive, then the critical point is a local minimum.

If $\lambda_1,\lambda_2$ and $\lambda_3$ are all negative, then the critical point is a local maximum.

Otherwise the critical point is a saddle point.

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Using principle minors

Let $|\textbf A_k|$ the k-th principle minor.

If $|\textbf A_k| \ \ \forall \ k$ are strictly postitive then the critical point is a minimum.

If $|\textbf A_{2k}| \ \ \forall \ k$ are strictly positive and if $|\textbf A_{2k+1}| \ \ \forall \ k$ are strictly negative then the critical point is a maximum.

Otherwise the critical point is a saddle point.

Here we have $|\textbf A_1|=|0|=0$. The critical points are saddle points.

Answer 2

Set $f_x,f_y,f_z$ all equal to 0 and solve for all the critical points. In this case, $y=-3$, $z=\pm\sqrt2$, $x=6\pm\sqrt2$. You should actually plug in your critical point one by one into the Hessian matrix (in this problem just plug in z), then find eigenvalues to check what type of extreme values they are.

Copied from Wikipedia:

For a function of more variables, one can look at the eigenvalues of the Hessian matrix at the critical point. The following test can be applied at a non-degenerate critical point (a, b, ...). If the Hessian is positive definite (equivalently, has all eigenvalues positive) at (a, b, ...), then f attains a local minimum at (a, b, ...). If the Hessian is negative definite (equivalently, has all eigenvalues negative) at (a, b, ...), then f attains a local maximum at (a, b, ...). If the Hessian has both positive and negative eigenvalues then (a, b, ...) is a saddle point for f (this is true even if (a, b, ...) is degenerate).