# Let \begin{pmatrix} 1 &2 \\ 2&1 \end{pmatrix}. Find the eigenvalues of A. If A is the Hessian...

## Question:

Let {eq}\begin{pmatrix} 1 &2 \\ 2&1 \end{pmatrix} {/eq}

Find the eigenvalues of A. If A is the Hessian matrix for f: {eq}\mathbb{R}^2\rightarrow \mathbb{R} {/eq} at a critical point in the domain of f, use the eigenvalues to classify the critical point.

## Eigenvalues and Critical Points:

We can classify the critical points of the function from the eigenvalues of the Hessian matrix. If the eigenvalues are of different signs, we can affirm that a local extreme at the critical point is not reached.

Given the matrix, {eq}A = \left( {\begin{array}{*{20}{c}} 1&2\\ 2&1 \end{array}} \right) {/eq}, we can calculate the eigenvalues of the matrix using determinant:

{eq}\left| {A - \lambda I} \right| = \left| {\begin{array}{*{20}{c}} {1 - \lambda }&2\\ 2&{1 - \lambda } \end{array}} \right| = \left( {1 - \lambda } \right)\left( {1 - \lambda } \right) - 4\\ = 1 - \lambda - \lambda + {\lambda ^2} - 4\\ = {\lambda ^2} - 2\lambda - 3 = 0\\ \lambda = \frac{{2 \pm \sqrt {{{\left( { - 2} \right)}^2} - 4 \cdot 1 \cdot \left( { - 3} \right)} }}{{2 \cdot 1}} = \frac{{2 \pm \sqrt {16} }}{2} = \frac{{2 \pm 4}}{2} = \left\{ {\begin{array}{*{20}{c}} {{\lambda _1} = 3}\\ {{\lambda _2} = - 1} \end{array}} \right. {/eq}

Finally, as the result are of different sign, we can say we have a saddle point. 