This is what my textbook wants me to do:
The matrix of the linear transformation $P_L$ that projects $\mathbb{R}^2$ on de straight line $l \leftrightarrow y = mx$ is:
\begin{pmatrix}
\frac{1}{1+m^2} & \frac{m}{1+m^2} \\
\frac{m}{1+m^2} & \frac{m^2}{1+m^2} \\
\end{pmatrix}
And I get this picture, which probably is their to inspire me:
Now, I tried to tackle this with the perpendicular line between A and C, but I got nowhere:
\begin{align*}
A &= (x_A, y_A) \\
&\Downarrow \\
AC \leftrightarrow y – y_A &= \frac{-1}{m}(x – x_A) \\
&\Downarrow \\
C &\leftrightarrow \begin{cases}
y = \frac{-x}{m} + \frac{x_A}{m} + y_A \\
y = mx \\
\end{cases}
\end{align*}
But that's were I am stuck, I can't get it to the form
\begin{align*}
\begin{bmatrix}
x_C \\
y_C
\end{bmatrix} &= \begin{bmatrix}
\frac{1}{1+m^2} & \frac{m}{1+m^2} \\
\frac{m}{1+m^2} & \frac{m^2}{1+m^2} \\
\end{bmatrix}
\begin{bmatrix}
x_A \\
y_A
\end{bmatrix}
\end{align*}
I am probably not doing it right. All tips are greatly appreciated.
Best Answer
You are off to a great start!
If you substitute your second equation into your first, you find $$mx=-\frac{x}{m}+\frac{x_A}{m}+y_A,$$ so $$\frac{1+m^2}{m}x=\frac{x_A}{m}+y_A,$$ and so $$x=\frac{1}{1+m^2}x_A+\frac{m}{1+m^2}y_A,\tag{$\star$}$$ which corresponds to what your first row should be.
Once you've found that, use $(\star)$ to substitute into your second equation, and you readily see that $$y=\frac{m}{1+m^2}x_A+\frac{m^2}{1+m^2}y_A,$$ which corresponds to the second row.