Let $V$ be the real vector space of $3 \times 3$ matrices with the bilinear form $\langle A,B \rangle=$ trace $A^tB$, and let $W$ be the subspace of skew-symmetric matrices. Compute the orthogonal projection to $W$ with respect to this form, of the matrix $$\begin{pmatrix} 1& 2 & 0\\ 0 & 0 & 1\\ 1 & 3 & 0\end{pmatrix}$$
Could someone show me how to proceed ?
On
Hint If $A$ is symmetric and $B$ is skew-symmetric, then $$ \mathrm{tr} (A^TB)=\mathrm{tr} (A^TB)^T=\mathrm{tr} (B^TA)=\mathrm{tr} (AB^T)=-\mathrm{tr} (A^TB)\quad \Rightarrow \quad \mathrm{tr} (A^TB)=0 $$
Hence
The orthogonal of subspace of skew-symmetric matrices is the subspace of symmetric matrices. It is easy to project onto the former parrallel to the latter. Just recall that $M=\frac{M+M^T}{2}+\frac{M-M^T}{2}$.
Therefore
The projection you are looking for is just $$\frac{M-M^T}{2}=\frac{1}{2}\pmatrix{ 1& 2 & 0\\ 0 & 0 & 1\\ 1 & 3 & 0}-\frac{1}{2}\pmatrix{ 1& 2 & 0\\ 0 & 0 & 1\\ 1 & 3 & 0}^T =\pmatrix{ 0& 1 & -1/2\\ -1 & 0 & -1\\ 1/2 & 1 & 0}$$
Find the orthogonal complement to $W$, i.e. $$W^{\perp} := \{ X \in V : \langle X, W \rangle = 0\}$$
Assuming that $\dim W + \dim W^{\top} = \dim V$, we can write any $X \in V$ as a linear combination:
$$X = \alpha A + \beta B$$ where $\alpha,\beta \in \mathbb{R}$, $A \in W$ and $B \in W^{\top}$. The orthogonal projection $\pi : V \twoheadrightarrow W$ is given by $$\pi : \alpha A + \beta B \longmapsto \alpha A$$