Let $v_{0}=\begin{bmatrix}0 & 1\\\ 1 & 0\end{bmatrix}$, prove that there is exactly one $h_0∈H$ having the property that $‖v_0-h_0‖≤‖v_0-h‖$ for all $h∈H$
we note that $H=span \begin{Bmatrix} \begin{bmatrix} 1 & 1\\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1\\\ 0 & 0 \end{bmatrix} \end{Bmatrix} $ is a subspace of $V=M_{2\times2}(\mathbb{R})$ which is an inner product space where $\langle a,b\rangle$ is defined to be the Frobenius Inner Product
the best guess that I could take is that we can assume $h_0$ is a vector such that $‖v_0-h_0‖$ is a minimum. Then using Caucy-Schwarz Inequality, we let $f(x) = ‖v_0-xh_0‖$ and prove that it is a local minima.
Is this correct? If so, how can I go on.