# The geometric action of an orthogonal $3 \times 3$ matrix with determinant $-1$.

I am trying to prove that the action of an orthogonal $3\times 3$ matrix with determinant $-1$ is a reflection about an eigenvector in one of the matrix's eigenspace, but I am a little lost.

Problem Statement: Describe geometrically the action of an orthogonal $3×3$ matrix with determinant $-1$.

What I have so far for the proof:

Claim that $A$ is an orthogonal $3×3$ matrix with determinant $-1$. Since $A$ is $3×3$, $A$ must have $3$ eigenvalues, $λ_1$, $λ_2$, and $λ_3$, and we know that $λ_1 λ_2 λ_3=-1$ must hold. Then we know that either

Case 1: $λ_1,λ_2,λ_3=-1$

Case 2: $λ_1,λ_2=1, λ_3=-1$

Case 3: $λ_1=-1, λ_2=c, λ_3=\bar{c}$ where $c,\bar{c}∈C$ and $\bar{c}$ is the complex conjugate of $c$.

Now, I am unsure what to do from here. My teacher was doing a similar proof in class where he was proving that the geometric action of a $3\times 3$ matrix with determinant $1$ is a rotation about one eigenspace. And based on his proof, I was thinking that I should do the following for this proof:

Let $v$ be an eigenvector with eigenvalue $-1$. Then we have $v^{⊥}=\left\{\ w∈F^{3} | v⋅w=0\right\}$ is a two dimensional vector subspace. Since $A$ is orthogonal, $Av^⊥⊂v^⊥$.

Now, based on what my teacher was telling me, he was saying the other two eigenspaces make a plane orthogonal to the eigenspace for $\lambda =-1$ and that $A$ reflects any vector about the eigenline for $\lambda = -1$?

Am I on the right track here? Any clarification for where I should go from here would be appreciated!

Here is a completely geometric description that doesn't have to talk about entries of A.

You've analyzed the eigenvalues correctly. Let v be an eigenvector for -1. Let T be the reflection of three space in the plane normal to v. Then TA=R is an orthogonal transformation with determinant 1, aka a rotation. In fact, you will easily see that v is an eigenvector for +1 for TA,=R so that it is the axis of rotation of TA (unless TA is already the identity, where it follows that T=A.)

Since T is its own inverse, you have the equation A=TR. Now you can see that A is a rotation around v, then a reflection through the plane normal to v.

$A$ reflects any vector about the eigenline for $\lambda = -1$

You mean rotates. In 3-space, you can rotate about a line or reflect in/through a plane, but "reflect about a line" is not sensical.

he was saying the other two eigenspaces make a plane orthogonal to the eigenspace for $\lambda =-1$.

The wrong part of this is that most of the time there are no other eigenspaces. You are guaranteed at least one dimension of eigenspace, but if the other two values are nonreal complex, there is no more eigenspace out there. The special cases are when the rotation is by 0 radians and by $\pi$ radians, where respectively there would be a two dimensional eigenspace for 1, or two more dimensions in the -1 eigenspace.

Here's a hint.

The restriction of the transformation $A$ to the orthogonal plane $v^\perp$ is itself the transformation of a $2 \times 2$ orthogonal matrix $B = \pmatrix{x & y \\ z & w}$ (with respect to an orthonormal basis $u_1,u_2$ of the 2-dimensional vector subspace $v^\perp$), such that the determinant of $B$ equals $1$. It follows that there is a $3 \times 3$ orthonormal basis change matrix $M$ such that $M v = \pmatrix{1 \\ 0 \\ 0}$, $M u_1 = \pmatrix{0 \\ 1 \\ 0}$, and $M u_2 = \pmatrix{0 \\ 1 \\ 1}$ and such that $$M A M^{-1} = \pmatrix{-1 & 0 & 0 \\ 0 & x & y \\ 0 & z & w}$$ So the $3 \times 3$ problem can be reduced to a $2 \times 2$ problem which you may already know: describe geometrically the action of an orthogonal $2 \times 2$ matrix with determinant $+1$.

• So I'm still a little confused. You are saying that $Av^{⊥}$ is equivalent to $Bw$ for some $2\times 2$ matrix $B$ and $w\in F^{2}$? And so if $v=<1,0,0>$ then the only vector in $v^{⊥}$ would be the zero vector? – yung_Pabs Oct 22 '15 at 2:04
• No, I'm talking about changing basis. I'll rewrite it more clearly. – Lee Mosher Oct 22 '15 at 12:51