Given was $v_1 = \begin{pmatrix} i\\0\\1 \end{pmatrix}$ and $v_2 = \begin{pmatrix} 0\\i\\1 \end{pmatrix}$.
1) I needed a orthonormalized basis $B$ of the sub-space built by $v_1, v_2$ and I got:
$B = (w_1, w_2) = ( \frac{1}{\sqrt{2}} \begin{pmatrix} i\\0\\1 \end{pmatrix}$, $ \frac{1}{\sqrt{6}} \begin{pmatrix} -i\\2i\\1 \end{pmatrix})$
2) I was told to add one vector so that I get an orthonormalized basis for $\mathbb{C}^3$:
I choose $w_3 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\i\\-i \end{pmatrix}$
3) I now need to determine the matrix for the orthogonalprojection onto subspace U relating to the orthonomalized basis B = (w_1, w_2, w_3)
What is this matrix of orthogonal projection and how can I calculate it?
If $U=\{v_1,v_2\}$, then the projection matrix is \begin{equation} A=[w_1,w_2,w_3]\begin{bmatrix}I_2 & 0 \\ 0&0 \end{bmatrix}[w_1,w_2,w_3]^*. \end{equation} This follows from the fact that for any $v \in U$, you will get $Av=v$, and if $w \notin U$ then $Aw=0$.