Let $f(x,y) = x^2-3xy+y^2$. Determine whether the point $(0,0)$ is a local maxima, local minima, or a saddle point using the eigenvalues of the Hessian of $f$ at the point $(0,0)$ or the eigenvalues of the associated symmetric matrix of $f$ .
This is start of my calculation:
The Hessian is a 2×2 matrix and looks like this:
$$
\text{Hess}(f) = \begin{bmatrix}
\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\
\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}
\end{bmatrix}
$$
Now, calculate the second derivatives of $f$:
$$
\frac{\partial^2 f}{\partial x^2} = 2
$$
$$
\frac{\partial^2 f}{\partial y^2} = 2
$$
$$
\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = -3
$$
So the Hessian of $f$ is:
$$
\text{Hess}(f) = \begin{bmatrix}
2 & -3 \\
-3 & 2
\end{bmatrix}
$$
Now we need to calculate the eigenvalues of this matrix. The eigenvalues can be calculated by solving the determinant equation:
$$
\text{det}(\text{Hess}(f) – \lambda E) = 0
$$
where $\lambda$ is the eigenvalue we are looking for and $E$ is the identity matrix. For our Hessian matrix we get:
$$
\text{det}\left(\begin{bmatrix}
2-\lambda & -3 \\
-3 & 2-\lambda
\end{bmatrix}\right) = 0
$$
How to continue?
Best Answer
Well, continuing from where you stopped, \begin{align*} \det \begin{pmatrix} 2 - \lambda & -3 \\ -3 & 2 - \lambda \end{pmatrix} &= 0 \\ (2-\lambda)(2-\lambda) - (-3)(-3) &= 0 \\ \lambda^2 - 4\lambda + 4 - 9 &= 0 \\ (\lambda - 5)(\lambda+1) &= 0. \end{align*} So the eigenvalues are $\lambda = 5, -1$.
What does this tell you about what type of stationary point $(0,0)$ is?
You can actually deduce the signs of the eigenvalues in the $2 \times 2$ case straight from the Hessian $H$. If the eigenvalues are $\lambda_1$ and $\lambda_2$, then here $\det H = \lambda_1 \lambda_2 = -5$ so the eigenvalues must have opposite signs.