Finding the matrix of projection Find the matrix A of the orthogonal projection onto the line L in R2 that consists of all scalar multiples of the vector $\begin{pmatrix} 2 \\ 3 \ \end{pmatrix}$. How do I begin to solve this? Any help would be appreciated.
 A: Hint:
The orthogonal projection of vector $v$ onto the mine directed by vector $u$ is given by
$$p_u(v)=\frac{\langle v,u\rangle}{\langle u,u\rangle}\,u$$
Hence all you have to do is to calculate the projections of the vectors of the basis onto the vector $\begin{pmatrix}2\\3\end{pmatrix}$.
A: Use the orthonormal basis for ${\sf R}^2$
$$\gamma=\{u_1,u_2\}=\left\{\frac{1}{\sqrt{13}}\begin{pmatrix}2\\3\end{pmatrix},\frac{1}{\sqrt{13}}\begin{pmatrix}3\\-2\end{pmatrix}\right\},$$
we see that
${\sf T}(u_1)=u_1$, ${\sf T}(u_2)=0$, and $[{\sf T}]_\gamma=\begin{pmatrix}1&0\\0&0\end{pmatrix}$. Let $\beta$ be the standard ordered basis for ${\sf R}^2$, and
write $Q=[{\sf I}]_\gamma^\beta=\frac{1}{\sqrt{13}}\begin{pmatrix}2&3\\3&-2\end{pmatrix}$. Then
\begin{align}
A=[{\sf T}]_\beta
&=Q[{\sf T}]_\gamma Q^{-1}
 =Q[{\sf T}]_\gamma Q^t
 =\frac{1}{13}\begin{pmatrix}2&3\\3&-2\end{pmatrix}
\begin{pmatrix}1&0\\0&0\end{pmatrix}
\begin{pmatrix}2&3\\3&-2\end{pmatrix}
=\frac{1}{13}\begin{pmatrix}4&6\\6&9\end{pmatrix}.
\end{align}
