Hessian matrix

2025-07-05

Imagine we got function and a few stationary points of this function, and we need to classify them, like: max, min or saddle point. If function without constraints, we can use Hessian matrix.

How to find stationary points of the function you can read at this article: “Stationary points and Lagrange method”

Let’s learn it on example. Function and stationary points:

$$ f(x, y) = x^3 + y^3 - 3xy; \quad A = (0; 0), \ B = (1; 1) $$

Firstly, build Hessian matrix. In general form Hessian matrix looks like:

$$ H_f(x) = \begin{bmatrix} \displaystyle \frac{\partial^2 f}{\partial x_1^2} & \displaystyle \frac{\partial^2 f}{\partial x_1 \partial x_2} & \cdots & \displaystyle \frac{\partial^2 f}{\partial x_1 \partial x_n} \\ \displaystyle \frac{\partial^2 f}{\partial x_2 \partial x_1} & \displaystyle \frac{\partial^2 f}{\partial x_2^2} & \cdots & \displaystyle \frac{\partial^2 f}{\partial x_2 \partial x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \displaystyle \frac{\partial^2 f}{\partial x_n \partial x_1} & \displaystyle \frac{\partial^2 f}{\partial x_n \partial x_2} & \cdots & \displaystyle \frac{\partial^2 f}{\partial x_n^2} \end{bmatrix} $$

For our function it is: $H = \begin{bmatrix}6x & -3 \\-3 & 6y\end{bmatrix}$. Then substitute points in this matrix and find eigenvalues:

$$ \begin{align} &H_A = \begin{bmatrix}0 & -3 \\ -3 & 0\end{bmatrix} &&H_B = \begin{bmatrix}6 & -3 \\ -3 & 6\end{bmatrix} \\[1em] &\det(H_A - \lambda I) = \det \begin{bmatrix} 0 - \lambda & -3 \\ -3 & 0 - \lambda \end{bmatrix} = &&\det(H_B - \lambda I) = \det \begin{bmatrix} 6 - \lambda & -3 \\ -3 & 6 - \lambda \end{bmatrix} = \\ &= (-\lambda)(-\lambda) - (-3)(-3) = \lambda^2 - 9 = 0 \ \Rightarrow \quad \quad \quad \quad &&= (6-\lambda)^2 - 9 = 0 \ \Rightarrow \\[1em] &\Rightarrow \ \lambda^2 = 9 \ \Rightarrow \ \lambda = \pm 3 \ \Rightarrow &&\Rightarrow \ (6-\lambda)^2 = 9 \ \Rightarrow \ 6 - \lambda = \pm 3 \ \Rightarrow \\[1em] &\Rightarrow \lambda_1 = 3, \quad \lambda_2 = -3 &&\Rightarrow \lambda_1 = 6 - 3 = 3, \quad \lambda_2 = 6 + 3 = 9 \end{align} $$

Then use this rules:

If Hessian eigenvalues is positives, point is local minimum
If Hessian eigenvalues is negative, point is local maximum
If Hessian eigenvalues not defined, point is saddle

As result, point $A$ it is saddle point and point $B$ it is local minimum.

Thanks for reading!

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