Determine representation matrix of orthogonal projection

Question

Let $V=\mathbb C^2$, $\mathcal E$ be the standard basis of $V$ and let $$U = \left\{ \begin{pmatrix} z_1 \\ z_2 \end{pmatrix}\in V | (1+i)z_1 + (3+2i)z_2 = 0 \right\}.$$ Finally let $p:V\to V$ be the orthogonal projection from $V$ to $U$ wrt the canoical scalar product on $V$. I'm trying to find that $A:=M_{\mathcal E}^{\mathcal E}(p)$, and I really do not know how. I could imagine that A is hermitian, or that I have to somehow construct a basis of U. How do I go about doing this? This has given me many troubles so far.

Answer 1

Note: Since the problem involves complex numbers, I assume everywhere you say "orthogonal" you mean "unitary." In this case the canonical scalar product between two vectors $\vec{v}, \vec{w}$ is $\vec{v}^* \vec{w}$, where the asterisk denotes the complex conjugate transpose of a matrix.

$U$ is one-dimensional (i.e. all vectors in it point in the same direction), so finding an orthonormal basis for it is easy. Note that any vector of $U$ is of the form $$\begin{pmatrix} z_1 \\ z_2 \end{pmatrix} = t\begin{pmatrix} 3 + 2i \\ -1 - i \end{pmatrix}$$ for some $t \in \mathbb{C}$. This is obtained just by solving the equation that vectors in $U$ satisfy. To find an orthonormal basis for $U$, we must find a vector of unit norm in $U$; this occurs when $$t = \left(|3 + 2i|^2 + |-1 - i|^2\right)^{-1/2} = \frac{1}{\sqrt{15}}$$ From here, let us denote $$\vec{u} = \frac{1}{\sqrt 15}\begin{pmatrix} 3 + 2i \\ -1 - i \end{pmatrix}$$ which is the sole element of an orthonormal basis for $U$. The orthogonal projection of some vector $\vec{v} \in V$ onto $U$ is then given by $$\text{proj}_{\vec{u}}(\vec{v}) = (\vec{v} \cdot \vec{u}) \vec{u} = \vec{u} \vec{u}^* \vec{v}$$ And hence we can see that the matrix representing the projection is given by $$A = \vec{u} \vec{u}^* = \frac{1}{15} \begin{pmatrix} 13 & 1 + 5i \\ 1 - 5i & 2\end{pmatrix}$$