Standard Matrix for an orthogonal projection transformation

[jmarzouq]

Let T:R^2 -> R^2 be the linear transformation that projects an R^2 vector (x,y) orthogonally onto (-2,4). Find the standard matrix for T.

I understand how to find a standard transformation matrix, I just don't really know what it's asking for. Is the transformation just (x-2, y+4)? Any clarification would be greatly appreciated, thanks! Also, sorry for the formatting, I'm on my phone.

 

[haruspex]

Can you define an orthogonal projection? If (-2, 4) is in the range, what would a vector of its null space look like?

 

[HallsofIvy]

You can do this pretty much by "brute strength". Since these vectors are in R², your matrix is 2 by 2 and looks like
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix}$$
and, because this is a projection onto [-2, 4], maps [-2, 4] to itself:
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}-2 \\ 4\end{bmatrix}= \begin{bmatrix}-2a+ 4b \\ -2c+ 4d\end{bmatrix}= \begin{bmatrix}-2 \\ 4\end{bmatrix}$$
so that -2a+ 4b= -2 and -2c+ 4p= 4. We can immediately divide by 2 to get -a+ 2b= -1 and -c+ 2d= 2.

Also, because this is an orthogonal projection, [4, 2], perpendicular to [-2, 4], must be mapped to 0.
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}4 \\ 2\end{bmatrix}= \begin{bmatrix}4a+ 2b \\ 4a+ 2d\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$$
so 4a+ 2b= 0 and 4c+ 2d= 0. We can again divide by 2 to get 2a+ b= 0 and 2c+ d= 0.

Solve the four equations -a+ 2b= -1, -c+ 2d= 2, 2a+ b= 0, and 2c+ d= 0 for a, b, c, and d.

 

[Robert1986]

Let's let $v$ be the normalised version of your vector (ie, divide your vector by its norm.) Now, if you have some vector $w$ and you want the projection onto $v$ then (as you probably know) this is just equal to $(v^\top w)v$ (since the norm of $v$ is 1.) But this is equal to $v(v^\top w)=(vv^\top)w$. Now, from this can you figure out how to form the projection?