# $3 \times 3$ Reflection Matrices

How can I tell if a $3 \times 3$ matrix is a reflection? I know that orthogonal matrices have determinants of either $1$ or $-1$. If the determinant is $-1$, then it can possibly be a reflection. How do I prove for sure that the $3 \times 3$ matrix is a reflection?

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A reflection matrix would either be a diagonal matrix with $\pm 1$ entries, or an elementary Hermitian (Householder) reflector... – J. M. Nov 3 '11 at 3:59

A square matrix $A$ of reals is a reflection if and only if $A$ is orthogonal ($A A^T = I$), also $A^2 = I$ and $I-A$ has rank one, where $I$ is the identity matrix. Checking rank can be done by the same techniques as Gaussian elimination, that is by row operations. At the end of the row operations, there is a single nonzero row remaining, call it the vector $\vec{v},$ and its transpose, a column vector, $\vec{v}^T.$ The action of $A$ on vectors (written as columns) is to send $\vec{v}^T$ to $-\vec{v}^T,$ and to leave alone any vector orthogonal to $\vec{v}.$ Note that we have said that $A$ is both orthogonal and symmetric, because we have said that $A=A^T.$

Examples are even possible with rational entries,

$$A \; = \; \frac{1}{49} \begin{pmatrix} 41 & 12 & -24 \\ 12 & 31 & 36 \\ -24 & 36 & -23 \\ \end{pmatrix}$$

Notice that

$$I-A \; = \; \frac{1}{49} \begin{pmatrix} 8 & -12 & 24 \\ -12 & 18 & -36 \\ 24 & -36 & 72 \\ \end{pmatrix}$$

When written as a row vector, we may take $\vec{v} = (2,-3,6)$ as the vector that is sent to its own negative. Note that the vectors $(-6,2,3)$ and $(3,6,2),$ which are orthogonal to each other and to $\vec{v},$ are fixed by $A.$

See:

http://mathoverflow.net/questions/73088/orthogonal-group-of-the-lattice-i-p-q/73116#73116

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If you are considering a matrix, you can find its eigenvectors, and thus diagonalize orthogonally if your matrix is assumed orthogonal. A reflection would have an eigenvalue of $1$ with a corresponding eigenspace of dimension $2$, plus an eigenvalue of $-1$ with a corresponding eigenspace of dimension $1$. This means that with respect to this basis, the reflection matrix looks like $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix}$$ thus it has determinant $-1$ since the matrices that change coordinates are not changing the determinant (you multiply by that matrix and its inverse so their determinants cancel out).

Now if you are considering a matrix and want to know if it is a reflection knowing that it has determinant $-1$, well first you have to assume it is orthogonal, because reflections are. Now assuming this, by diagonalizing it orthogonally, the eigenvalues MUST be $1$ and $-1$, and since you have determinant $-1$, the eigenvalue $1$ will have a 2-dimensional eigenspace and $-1$ will have a 1-dimensional eigenspace. If the eigenvalues are NOT $1$ and $-1$, then it cannot be a reflection in the strict sense of the term (i.e. it might stretch the vectors in some direction instead of just reflecting).

Hope that helps,

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The "inversion", diag(-1,-1,-1) looks a lot like a reflection (orthogonal, determinant -1, and does not stretch any vectors). I'm not sure of your description of the the "fake reflections" (once we assume they are orthogonal and determinant -1, I think you get reflections, the inversion, and rotary reflections). – Jack Schmidt Nov 3 '11 at 4:12
I meant an "inversion" in the sense of inverting with respect to an axis. And note that the matrix diag(-1,-1,-1) does not have eigenvalues 1 and -1, and does stretch the vectors in the "opposite direction". – Patrick Da Silva Nov 3 '11 at 4:14
Also note that rotary reflections do not have $1$ as an eigenvalue. – Patrick Da Silva Nov 3 '11 at 4:16
Oh yes, sorry I should be clear: yes, your description of needing 1 and -1 as eigenvalues is spot on (hence the +1). It was only your description of the "not having a 1" guys that sounded fishy. It might just be a matter of language though, if you think of an eigenvalue of √−1 as stretching them around to the left or something. – Jack Schmidt Nov 3 '11 at 4:19