Having just finished a post on eigenvalue decomposition, involving eigenvalues and eigenvectors, I was flipping through this nifty little textbook, where I saw a description of geometric reflections using eigenvectors, and I thought, “Oh my gosh, of course!”

We have seen reflections here, yet back then we had to find something called the foot of a point to figure out the reflection. But we can construct a reflection matrix (same as a scaling matrix) using only knowledge about eigenvectors.

When we talked about eigenvectors before—those vectors which do not change direction under a transformation (except if they reverse direction)—we were looking for them. But in a reflection, we already know them. In any reflection across a line, we already know that the vector that matches the line of reflection will not change direction, and the vector perpendicular to the line of reflection will only be scaled by \(\mathtt{-1}\). So, both the vector which describes the line of reflection and the vector perpendicular to that line are eigenvectors.

So let’s say we want to reflect point \(\mathtt{C}\) across the line described by the vector \(\mathtt{\alpha(2, -1)}\).

Our eigenvector matrix will be \[\begin{bmatrix}\mathtt{\,\,\,\,2}&\mathtt{-1}\\\mathtt{-1}&\mathtt{-2}\end{bmatrix}\] Here we see the second eigenvector, but we can simply use the vector perpendicular to the first if we don’t know the second one.

Our eigenvalue matrix will be \[\begin{bmatrix}\mathtt{1}&\mathtt{\,\,\,\,0}\\\mathtt{0}&\mathtt{-1}\end{bmatrix}\] since we will keep the first eigenvector—the line of reflection—fixed (eigenvalue of 1) and just flip the second eigenvector (eigenvalue of –1).

As a penultimate step, we calculate the inverse of the eigenvector matrix (we’ll get into inverses fairly soon), and then finally multiply all those matrices together (right to left), as we saw with the eigenvalue decomposition, to get the reflection matrix. \[\begin{bmatrix}\mathtt{\,\,\,\,2}&\mathtt{-1}\\\mathtt{-1}&\mathtt{-2}\end{bmatrix}\begin{bmatrix}\mathtt{1}&\mathtt{\,\,\,\,0}\\\mathtt{0}&\mathtt{-1}\end{bmatrix}\begin{bmatrix}\mathtt{\,\,\,\,\frac{2}{5}}&\mathtt{-\frac{1}{5}}\\\mathtt{-\frac{1}{5}}&\mathtt{-\frac{2}{5}}\end{bmatrix}\mathtt{=}\begin{bmatrix}\mathtt{\,\,\,\,\frac{3}{5}}&\mathtt{-\frac{4}{5}}\\\mathtt{-\frac{4}{5}}&\mathtt{-\frac{3}{5}}\end{bmatrix}\]

The final matrix on the right side is the reflection matrix for reflections across the line represented by the vector \(\mathtt{\alpha(2, -1)}\), or essentially all lines with a slope of \(\mathtt{-\frac{1}{2}}\).

It’s worth playing around with building these matrices and using them to find reflections. A lot of what’s here is pretty obvious, but reflection matrices can come in handy in some non-obvious ways too. The perpendicular vector at the very least has to be on the same side of the line as the point you are reflecting.