0

I've been asked to find a matrix representing projection of a point in the plane onto the line $ax + by = 0.$

I know the matrix for $ax+by$ is $$\left[\begin{array}{cc}a &0\\ b&0\end{array}\right]$$ I also know projection of $u$ onto $v$ is $((u\cdot v)/\|v\|^2)v$. What would I be projecting onto?

A second question is finding a matrix for reflection about the line $4x + 5y = 0$ followed by rotation by $\pi/2$ radians clockwise about the origin. Matrix: $$\left[\begin{array}{cc}4& 0\\ 5& 0\end{array}\right].$$ Where should I go from here?

  • The matrices you are writing down have nothing to do with the problem. Instead, find a vector on the line. That's the vector you are trying to project onto. – Gerry Myerson Oct 16 '17 at 05:32
  • So how should I proceed? –  Oct 16 '17 at 22:07
  • As I said, start by finding a (nonzero) vector on the line. You have the line $ax+by=0$. Find a vector (which is to say, a point) on that line. Can you do that? – Gerry Myerson Oct 16 '17 at 22:19
  • Is vector the same as vector equation? –  Oct 16 '17 at 22:21
  • I think I forgot how to do it –  Oct 16 '17 at 22:23
  • You forgot how to find a point on a line, given an equation for the line? OK. You have a line with equation $ax+by=0$. What's your favorite number (other than zero)? Let $x$ be that number. Now: what value do you get for $y$? – Gerry Myerson Oct 16 '17 at 22:31
  • I let x=3, y=-3a/b –  Oct 16 '17 at 22:43
  • where do i go next? –  Oct 16 '17 at 22:51
  • OK, so that gives you a point, $v=(3,-3a/b)$, on the line (provided $b\ne0$, but you can deal with that case later). So that $v$ is what you are projecting onto. And you know the formula for projecting $u$ onto $v$. So if you let $u$ be a general point in the plane, $u=(x,y)$, you can use that formula and you can compute the projection of $u$ onto $v$. Right? – Gerry Myerson Oct 16 '17 at 23:02
  • so my u would be (x, y) and my v would be (3,-3a/b)? –  Oct 16 '17 at 23:22
  • doing that i get p(u onto v)=(9x,9a^2yb)/(9+(9a^2)/b^2) –  Oct 16 '17 at 23:25
  • where should I proceed after this? –  Oct 17 '17 at 01:34
  • I yield to Tyler6. – Gerry Myerson Oct 17 '17 at 04:01

1 Answers1

1

For writing the matrix of any transformation, the columns $[a_1 a_2 ... a_n]$ are where the unit vectors $e_1,e_2,...e_n$ are mapped to by the transformation. For your first question, we are projecting onto the line $ax+by=0$,which is also the span of the vector $$\begin{pmatrix}-\frac{1}{a}\\\frac{1}{b}\end{pmatrix}$$ So, to find the transformation we calculate the image of $e_1$ and $e_2$ when they are projected onto this span, and take those new coordinates to be the columns of the matrix.

For the second one, it is slightly more complicated. Note that the projection of a vector onto a span is halfway between the original vector and the reflected vector. Letting $\vec P$, $\vec R$, $\vec x$ be the projected vector, reflected vector and original vector respectively, we have $ \vec P =\frac{1}{2}(\vec R +\vec x)$, or $\vec R=2\vec P-\vec X.$ This can be translated into matrices by taking the matrix of the projection for $\vec P$, the matrix of the reflection for $\vec R$, and the identity matrix for $\vec X$.This gives us $[R]=2[P]-[I]$.

Applying this to $e_1$ and $e_2$ will give the columns for our reflection matrix, and now we apply the rotation matrix, $$\begin{pmatrix}\cos(\theta) &-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{pmatrix}$$ where $\theta$ is the angle of rotation. You will plug in not $e_1$ and $e_2$, but their images under the initial transformation, and their new images under this transformation will be the columns of your new matrix.

Note: For the second part, an alternate method would be to find the matrix of the reflection and rotation individually, and then use matrix multiplication as it corresponds to the composition of functions.

Tyler6
  • 1,219
  • 10
  • 24