Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

Projection of A to C on a line

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.

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

You are off to a great start! (You have a typo in the matrix, though. Each denominator should be $1+m^2$.)

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 $$x=\frac{1}{1+m^2}x_A+\frac{m}{1+m^2}y_A,$$ which corresponds to what your first row should be.

Once you've found that, sub that $x$ expression 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.

share|improve this answer
    
So, I was close! Thank you very much for the last missing bits. It seems obvious now. –  hver Jun 17 '12 at 19:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.