Why has the kernel of the 2x2 hessian 3 linear functions?

Question

I have calculated something characteristic polynome and such things but i think thats the wrong way to see this. My try is :

On $\mathbb{R}^2$ the Hessian matrix can be relaized as $$ \mathbf{H}_\mathbb{R^2}= \begin{pmatrix} \partial_{xx} & \partial_{xy} \\ \partial_{yx} & \partial_{yy} \end{pmatrix} \Leftrightarrow \begin{pmatrix} \partial_{xx} & \partial_{xy} \\ \partial_{xy} & \partial_{yy} \end{pmatrix} $$ The kernel, or null space, of a matrix is defined as all vectors that are mapped to the zero vector and the kernel is not trivial if the kernel not only contains the zero element. For the Hessian matrix then the kernel is the set of functions that have all second partial derivatives mapped to zero. The only three functions for that this holds are $u(x,y) = x$, $u(x,y) = y$ and $u(x,y) = x + y$, so $\dim(\ker(\mathbf{H}_u)) = 3$.

Is this correct ?

Answer 1

No, because the third function is a linear combination of the first and the second. And they are not the only ones. All their linear combinations belong to kernel.

The dimension is $3$, by the way. The kernel is the set $$ \ker \mathbf{H} = \{\lambda x + \mu y + \varepsilon 1 : \lambda, \mu, \varepsilon \in \mathbb{R}\}. $$