# Matrix for the reflection over the null space of a matrix

I'm working on my latest linear algebra assignment and one question is as follows:

In $\mathbb R^3$ let R be the reflection over the null space of the matrix

A = [4 4 5]

Find the matrix which represents R using standard coordinates.

I am familiar with the fact that the matrix for the orthogonal projection onto a subspace is given by [P] = A($A^TA$)$A^T$, where A is a matrix with columns that form a basis for the space in question. However, I am not familiar with the concept of a "reflection over the null space". How does this compare with the matrix for the orthogonal projection onto the null space? Is it a similar process? Thanks for any help.

First of all, the formula should be $$P = B(B^TB)^{-1}B^T$$ where the columns of $B$ form of a basis of $ker(A)$.

Think geometrically when solving it. Points are to be reflected in a plane which is the kernel of $A$ (see third item):

• find a basis $v_1, v_2$ in $ker(A)$ and set up $B = (v_1 \, v_2)$
• build the projector $P$ onto $ker(A)$ with above formula
• geometrically the following happens to a point $x = (x_1 \, x_2 \, x_3)$ while reflecting in the plane $ker(A)$: $x$ is split into two parts - its projection onto the plane and the corresponding orthogonal part of $x$. Then flip the direction of this orthogonal part: $$x = Px + (x - Px) \mapsto Px - (x-Px) \rightarrow x \mapsto Px - (I-P)x = (2P-I)x$$ So, the matrix looked for is $$2P-I$$
• Ah, you beat me to it! I think I will leave my answer as well because it gives a slightly different perspective on the same idea, but you were first. +1 – RCT Mar 25 '18 at 3:20

I could imagine that there's some ugly formula for this, but I don't know it, and it's probably more instructive to solve the problem from basic principles anyway, so let's try that! My first thought when reading this problem is to start by changing to a basis in which it's obvious what the reflection does.

Note that multiplication by $A$ is just dotting with the vector $(4,4,5),$ so the kernel of $A$ is precisely $(4,4,5)^\perp.$ So $(4,4,5)$ is orthogonal to the plane of reflection, and thus the reflection simply negates it.

Now extend $\{(4,4,5)\}$ to a basis for $\mathbb{R}^3$ by picking a basis for the null space itself. For example, $(0,-5,4)$ and $(-5,0,4)$ are two independent vectors killed by $A.$ Since these live in the plane of reflection, they are fixed by the reflection.

So in our new basis, the matrix for the reflection is simply $$R' = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$

Now to write the matrix in the standard basis, all you have to do is compute the inverse of the change of basis matrix $$P = \begin{bmatrix} 4 & 0 & -5 \\ 4 & -5 & 0 \\ 5 & 4 & 4 \end{bmatrix}$$ that took the standard basis to our preferred basis. It ends up being $$P^{-1} = \frac{1}{285}\begin{bmatrix} 20 & 20 & 25 \\ 16 & -41 & 20 \\ -41 & 16 & 20 \end{bmatrix}.$$

So $R$ is obtained by changing to the new basis, applying $R',$ then changing back to the standard basis, i.e. $R = P^{-1}R'P.$