Linear Algebra problem here. I was given the equation of a line and told to find a matrix for it; I found the matrix for orthogonal projection... not reflection.

Here's the problem:

Let W be the line $x = 2t$; $y = -t$; $z = 4t$; $w = -3t$ in $\mathbb{R}^4$. Find the matrix for orthogonal reflection on W in the standard basis.

When finding the projection matrix, I just created a vector out of the equation for the line, normalized it, then multiplied it by itself to produce the $4\times 4$ matrix. This worked, but I don't know a way (if one exists) to convert that matrix into one for reflection instead of projection.

I know that the first column of the matrix is based off the equations for the line, so the first vector is $(2, -1, 4, -3)$. And I know that I use the equations to find the other three vectors for the matrix. Each vector $(x, y, z, w)$ has to be such that:

$2x - y + 4z - 3w = 0$.

But I can't just pick any number and find the other values. How do I use this equation in a way that will produce vectors that form a basis for reflection and not just a projection?

  • 1
    $\begingroup$ What’s the matrix of a line? $\endgroup$ – amd May 1 '18 at 20:34
  • $\begingroup$ Welcome to Math.SE! Please use MathJax. For some basic information about writing math at this site see e.g. basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 1 '18 at 20:35
  • $\begingroup$ Think about how reflection in a line relates to projection onto it. What happens to the component of the vector that’s orthogonal to the line? $\endgroup$ – amd May 1 '18 at 20:35
  • $\begingroup$ The vector component that's orthogonal doesn't change, and if it's reflection it would be reversed, but I'm pretty sure I can't just flip some signs in the orthogonal projection matrix. $\endgroup$ – Sabrina May 1 '18 at 20:56
  • $\begingroup$ Good. Try writing what you’ve just said out as an equation and see if you can factor out the vector being reflected. $\endgroup$ – amd May 1 '18 at 22:22

You’re very close. You’ve already worked out $P_W$, the orthogonal projection onto $W$, and know that a reflection reverses the component of a vector orthogonal to the reflector. With that in hand, you can construct the reflection matrix. Let $\mathbf v_\parallel = P_W\mathbf v$ and $\mathbf v_\perp = \mathbf v-\mathbf v_\parallel = \mathbf v-P_W\mathbf v$. Then the reflection of $\mathbf v$ in $W$ is $$R\mathbf v = \mathbf v_\parallel - \mathbf v_\perp = \mathbf v_\parallel - (\mathbf v - \mathbf v_\parallel) = 2\mathbf v_\parallel - \mathbf v = 2P_W\mathbf v-\mathbf v = (2P_W-I)\mathbf v.$$ Observe that we haven’t made any assumptions about the subspace $W$. This construction works whether $W$ is a line, plane, &c.

The approach you mention toward the end of your question has some good ideas, but it looks like you might be getting some of them mixed up. The columns of a transformation matrix are the images of the basis vectors. Unfortunately, this means that the first column of the reflection matrix is not $\mathbf w = (2,-1,4,-3)^T$, the direction vector of the line. If it were, that would mean that the standard basis vector $(1,0,0,0)^T$ gets mapped onto the line, which isn’t at all what you want. On the other hand, if you take $\mathbf w$ as your first basis vector, then relative to that basis the first column of the reflection matrix is $(1,0,0,0)^T$. As stated previously, the reflection reverses orthogonal rejections from $W$, so to continue this construction, find a basis for $W^\perp$ and adjoin it to $\mathbf w$. Relative to this basis, the reflection matrix is simply $\operatorname{diag}(1,-1,-1,-1)$ and a change of basis back to the standard basis will give you the required reflection matrix. All that remains is to find three linearly-independent vectors that are all orthogonal to $\mathbf w$, which can be done by inspection.

  • $\begingroup$ I don't understand what you mean by the vectors you list. (1,0,0,0) can't be in the reflection matrix because it isn't orthogonal to (2,-1,4,-3), which you say should be used as a basis instead of a vector in the matrix. $\endgroup$ – Sabrina May 2 '18 at 20:53
  • $\begingroup$ @Sabrina You build the reflection matrix in a basis different from the standard basis, one in which it has a very simple form. Relative to a coordinate system in which $(2,-1,4,-3)^T$ is the first basis vector, its own coordinates are $(1,0,0,0)^T$ and it gets mapped to itself by the reflection, so relative to that basis the first column of the matrix is $(1,0,0,0)^T$. $\endgroup$ – amd May 2 '18 at 21:26
  • $\begingroup$ What does it mean for the coordinates to be (1,0,0,0) relative to that basis? $\endgroup$ – Sabrina May 3 '18 at 0:42

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