Finding the matrix of linear transformation What is the orthogonal projection on the line of equation $x = y$ of the point
$\begin{pmatrix} 3 \\ -1 \end{pmatrix}$? Assume this is a linear transformation. 

The matrix for this linear transformation is $\begin{bmatrix} \frac 12 & \frac 12 \\ \frac 12& \frac 12 \end{bmatrix}$ since the perpendicular line from $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ to the line of equation $x = y$ intersects that line at $\begin{pmatrix} \frac 12 \\ \frac 12 \end{pmatrix}$, as does the perpendicular line from $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. To determine the orthogonal projection of the point $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$, we multiply $\begin{bmatrix} \frac 12 & \frac 12 \\ \frac 12& \frac 12 \end{bmatrix}$$\begin{bmatrix} 3 \\ -1 \end{bmatrix} =$$\begin{bmatrix} 1 \\ 1 \end{bmatrix}.$

I am having hard time seeing:

the perpendicular line from $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ to the line of equation $x = y$ intersects that line at $\begin{pmatrix} \frac 12 \\ \frac 12 \end{pmatrix}$

Wouldn't $x = y$ be intersected at $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$?

When your quote says "the perpendicular line from $(1,0)$ to the line of equation $x=y$", what is meant is the line through $(1,0)$ that is perpendicular to $x=y$.
It looks like you understood it as perpendicular to the vector $(1,0)$, but that is not how an orthogonal projection works.
