# Second Derivative Test and Hessian for $f(x,y) = x^2 + y^2$.

My task was to find the critical points of the function $f(x,y) = x^2+y^2$, to then compute the Hessian, and to use the second derivative test to determine whether the critical points are local maxima or minima.

I think I did it correctly, but I'm new to multivariate calculus, so I'd like someone to check my work.

Solution.

$f$ has a critical point when all the partial derivatives are $0$. The partials are $$\frac{\partial}{\partial x} (x^2+y^2) = 2x$$ and $$\frac{\partial}{\partial y} (x^2+y^2) = 2y$$ So $Df(x,y) = [2x \quad 2y]$. Clearly $2x=2y=0$ iff $x=y=0$. Now we need to compute the second-order partial derivatives: $$\frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}\right) = \frac{\partial}{\partial x} 2x = 2 = D_{xx}f(x,y)$$ $$\frac{\partial}{\partial x}\left(\frac{\partial}{\partial y}\right) = \frac{\partial}{\partial x} 2y = 0 = D_{yx}f(x,y)$$ $$\frac{\partial}{\partial y}\left(\frac{\partial}{\partial x}\right) = \frac{\partial}{\partial y} 2x = 0 = D_{xy}f(x,y)$$ $$\frac{\partial}{\partial y}\left(\frac{\partial}{\partial y}\right) = \frac{\partial}{\partial y} 2y = 2 = D_{yy}f(x,y)$$

The Hessian is then given by: $$\textbf{H}(x,y) = \left[ \begin{matrix} D_{xx}f(x,y) & D_{xy}f(x,y) \\ D_{yx}f(x,y) & D_{yy}f(x,y) \end{matrix}\right] = \left[ \begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}\right]$$ By the second derivative test, $(0,0)$ is a local minimum if $\textbf{H}(0,0)$ is positive definite, and a local maximum if $\textbf{H}(0,0)$ is negative definite. $\textbf{H}$ is positive definite if $z^T\textbf{H}z$ is positive for every non-zero column vector $z \in \Bbb R^n$. Thus, take $z = [a \quad b]$. We have that $$[a \quad b]\left[ \begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}\right]\left[\begin{matrix} a \\ b\end{matrix}\right] = [a \quad b] \left[ \begin{matrix} 2a + 0b \\ 0a + 2b \end{matrix} \right] = [a \quad b] \left[ \begin{matrix} 2a \\ 2b \end{matrix} \right] = [2a^2 + 2b^2]$$ Which is clearly positive definite since $x^2 \geq 0 \forall x\in\Bbb R$. So $(0,0)$ is a local minimum.

• When checking for positive/negative definiteness, it is more efficient (and more common, in this context) to do so using Sylvester's criterion – Omnomnomnom May 6 '15 at 12:13

Perfect answer, but a small comment with regard to the last matrix computation: $$\begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} 2a \\ 2b \end{bmatrix} = 2a^2+2b^2$$ Remember that the quadratic form $z^T H z$ is a real number, not a matrix. We call the graph of the function $z=x^2+y^2$ an elliptic paraboloid, btw.
• Why is it a real number (scalar), rather than a $1\times 1$ matrix? – Newb May 6 '15 at 12:03
• Ah, thanks. By the way, it appears there is some controversy regarding the equivalence of $1\times 1$ matrices and scalars: see math.stackexchange.com/questions/65002/… – Newb May 6 '15 at 12:08
• I don't know linear orders that respect multiplication on matrices, but I do know this for real numbers. Now, you can call this $1\times 1$ matrix, but what does positive matrix mean? As is pointed in the link you just gave, this is unnecessary notation overloading. There is a reason why we think of vectors and scalars separately. – Ennar May 6 '15 at 12:12