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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:

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

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$M=I-2P$. 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 $λ=±1$

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    $\begingroup$ @hardmath not sure why that is necessary… $\endgroup$ – YiFan Oct 7 at 21:57
  • $\begingroup$ @YiFan: For the few words added, see the edit history. It ties out the Answer with the Question as asked. $\endgroup$ – hardmath Oct 7 at 22:34
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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 $-1$.

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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 $-(+1)$.

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

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