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Show that if a $3\times 3$ matrix $A$ represents the reflection about a plane, then $A$ is similar to the matrix $\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1 \end{bmatrix}$

This is not homework, just me preparing for a test -- but how can this be solved? I know the definition of similarity, but can't find a way to approach the problem.

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I can't comment, but what if my $A$ matrix is $\begin{bmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$ (so reflection across yz plane)? Then the basis as you say could be formed via columns of the same matrix, since the first column is orthogonal to the yz plane, and the rest two are parallel. But $\begin{bmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$*$\begin{bmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$ (that is, $A$ times basis) does not equal to $\begin{bmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$ times $\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\ –  user6901 Feb 11 '11 at 10:39
    
Sorry for the formating, I meant $\begin{bmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$ times $\begin{bmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$ (which is $A$ times our basis which can be formed using columns of $A$) does not equal to $\begin{bmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$ times $\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1 \end{bmatrix}$ which is supposed to be the case if they are similar. –  user6902 Feb 11 '11 at 10:39
    
@Jason: the fact that you can't comment on answers to your own questions means you are not registered. Please register instead of putting comments in answers. –  Qiaochu Yuan Feb 11 '11 at 10:40
    
@Jason: Please don't use the subject line as part of the message. The body should be self-contained, and the subject should describe the kind of thing you are asking about. –  Arturo Magidin Feb 11 '11 at 14:32

1 Answer 1

Suppose that $A$ is a reflection about a plane spanned by $u,v$ and let $w$ be orthogonal to this plane, then $Au=u$ and $Av=v$ since the reflection through the plane doesn't change them and $Aw = -w$ (this is the reflection).

Now switch the basis to $u,v,w$ and you will get the result.


If you have any two diagonal matrices, that have the same entries on their diagonal, but maybe in a different order, then they are similar.

You can see this by defining the matrix $E_\sigma$ for a permutation $\sigma \in S_n$ on n elements as $(E_\sigma)_{i,\sigma(i)}=1$ and all the other entries are zero. If $D$ is diagonal then $E_\sigma D E_\sigma ^{-1}$ is just the permutation of the entries on the diagonal.

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