# How can I prove mathematically the reflection matrix has only the eigenvalues 1 or -1?

Specifically where S is a subspace of $$R^n$$. $$P_S$$ is the orthogonal projection onto $$S$$.

and the reflection matrix $$M = I - 2P_S$$

I understand a similar proof where the eigenvalues of the projection matrix is either 0 or 1. Now trying to get the intuition for the reflection matrix (M) case.

This is the proof for projection matrices that I have seen:

$$Px = \lambda x$$ $$P^2 = P$$ $$P^2x = \lambda x$$ $$P(Px) = \lambda x$$ $$\lambda^2x = \lambda x$$ $$\lambda(\lambda -1)x = 0$$

$$M=I-2P$$. Then $$M^2=(I-2P)(I-2P)=I^2-4P+4P^2=I-4P+4P=I$$ and $$M^2x=λ^2x=Ix=x$$. Then $$λ^2=1$$. Hence $$λ=±1$$

• @hardmath not sure why that is necessary… – YiFan Oct 7 at 21:57
• @YiFan: For the few words added, see the edit history. It ties out the Answer with the Question as asked. – hardmath Oct 7 at 22:34

Let $$\lambda_S\in$${$$0,1$$} denote the eigenvalue of projection matrix $$P_S$$. The given relation $$M=I-2P_S=f(P_S)$$ where $$f(P_S)$$ is matrix polynomial in $$P_S$$. If $$\lambda_M$$ is the eigenvalue of $$M$$ then $$\lambda_M=1-2\lambda_S=1-2(0)$$ or $$1-2(1)$$ i.e. $$1$$ or $$-1$$.

If you're reflecting in the subspace $$S$$ then $$S$$ is the $$1$$-eigenspace and $$S^\perp$$ is the $$(-1)$$-eigenspace. Since $$S \oplus S^\perp = \mathbf{R}^n$$ there are no additional possible eigenvalues.

This, to me, is the definition of reflection in the space $$S$$: it fixes $$S$$ and it sends a vector orthogonal to $$S$$ to its negative. To write this in terms of $$P_S$$ we use that $$P_S$$ is the identity operator on $$S$$ and $$1 - P_S$$ is the identity operator on $$S^\perp$$ (and they are zero on the complementary space). So the reflection map is $$P_S - (I - P_S)$$ (identity on $$S$$ + the negative identity on $$S^\perp$$). That gives you $$-I + 2P_S$$.

Now, for completeness, you can verify: if $$x \in S$$ (so $$P_S x = x$$) then $$(-I + 2P_S)x = -x + 2x = x$$ and if $$x \in S^\perp$$ (so $$P_S x = 0$$) then $$(-I + 2P_S)x = -x$$.

The matrix you gave I do not consider to be reflection in $$S$$. But, if my matrix has eigenvalues $$-1$$ and $$+1$$, then yours—being the negative of mine—has eigenvalues $$-(-1)$$ and $$-(+1)$$.

• Justification for these assertions would improve the answer. – Greg Martin Oct 7 at 23:46
• @Greg better now? – Trevor Gunn Oct 8 at 0:21

If $$M$$ is a reflection matrix, then $$M^2=I$$ (the identity matrix), so $$M^2-I=0$$. This means that $$p(x)=x^2-1$$ "cancels" $$M$$ (1). As a consequence, the minimal polynomial of $$M$$ is either $$\mu(x)=x-1$$ (then $$M=I$$) or $$\mu(x)=x+1$$ (then $$M=-I$$) or $$\mu(x)=x^2-1$$. In all cases, the zeroes of $$\mu$$, $$-1$$ and/or $$1$$, are the eigenvalues of $$M$$.

(1) I couldn't find the terminology for a polynomial $$p(x)$$ such that $$p(M)=0$$ (polynomial annulateur in French).