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I read earlier that if $A$ is a real $3\times 3$ matrix satsifying $A^3=I$, then $A$ is similar to a matrix of form $$ \begin{bmatrix} 1 & 0 & 0\\ 0 & \cos\theta & -\sin\theta\\ 0 & \sin\theta & \cos\theta \end{bmatrix} $$

From the structure theorems for normal operators, I know that among the real normal operators, the othogonal operators are exactly those matrices which are block diagonal with their eigenvalues and $2\times 2$ blocks of the above form.

So I suppose $A$ must be normal/orthogonal? Is there a way to deduce that at all from just knowing that $A^3=I$ without appealing to the structure theorem in hindsight?

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A is orthogonal iff it is orthogonally similar to an orthogonal matrix. Plain old similarity to an orthogonal matrix does not imply orthogonality. –  Eric Auld Sep 6 '13 at 3:37
    
Note that, $A^3=I \implies |A|^3=1$. –  Mhenni Benghorbal Sep 6 '13 at 3:59
    
The above matrices (and their orthogonal conjugates) describe the proper orthogonal matrices. –  copper.hat Sep 6 '13 at 8:19
    
@MhenniBenghorbal May I ask how you've reached that conclusion? I thought the matrix norm only gave us $\|AB\| \leq \|A\|\|B\|$, not $\|AB\| = \|A\|\|B\|$. –  Eric Auld Sep 6 '13 at 13:06
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3 Answers

This is not true. That $A^3=I$ implies that $A$ is diagonalizable over $\mathbf{C}$ and that its eigenvalues are cube roots of unity and thus on the unit circle.

And while being diagonalizable with eigenvalues on the unit circle is necessary for being unitary (over $\mathbf{R}$, orthogonal), it is not sufficient: the corresponding eigenvectors must be orthogonal for different eigenvalues.

A counterexample can be found simply by conjugating an orthogonal matrix with a matrix that is not orthogonal.

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I agree. +1--------- –  Eric Auld Sep 6 '13 at 3:30
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Let $A=\begin{bmatrix} 1 & -\frac{3}{2} & \frac{\sqrt{3}}{2} \\ 0 & -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ 0 & \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix}$. Then $A^3=I$, but $A$ is not orthogonally similar to the matrix above since the matrix above is normal whereas $A$ is not.

Elaboration:

To construct $A$ we note that we must have $\lambda^3 = 1$ for all eigenvalues $\lambda$, so we have two choices of eigenvalues (since if $\lambda$ is an eigenvalue, so must $\overline{\lambda}$): $\{1\}$ (multiplicity 3), or $\{1, e^{i 2 \pi/3}, e^{-i 2 \pi/3} \}$.

If the eigenvalues are all $1$, then we must have $A=I$, which is uninteresting. To see this, since all eigenvalues are 1, Cayley Hamilton gives $(A-I)^3 = 0$, which gives $A^2 = A$ (since $A^3 = I$), multiplying across by $A$ gives $A^3 = A^2 $ from which it follows that $A=I$.

So we have eigenvalues $1$, $\lambda = e^{i 2 \pi/3}$, and $\overline{\lambda}$. (Note that if $A v = \lambda v$, then $A \overline{v} = \overline{\lambda} \overline{ v}$.) To finish I want to pick three linearly independent, non-orthogonal eigenvectors. I choose $e_1$ for the eigenvalue $1$, and $(1,1,i)^T$ for the eigenvalue $\lambda$ (hence the eigenvector for the eigenvalue $\overline{\lambda}$ must be $(1,1,-i)^T$). Multiplying these out appropriately gives the above matrix.

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@Did: Thanks for catching that. I have repaired the solution. –  copper.hat Sep 6 '13 at 8:23
    
No magic really, just grunt work. I choose an eigenvector of $e_1$ for eigenvalue $1$, $(1,1,i)^T$ for $e^{i \frac{2 \pi}{3}}$, and the appropriate conjugates for the other one. Then I multiplied them out to get the above. –  copper.hat Sep 6 '13 at 8:30
    
@Did: I added all the detail I had on my scratch paper, I'm not sure it is interesting though. –  copper.hat Sep 6 '13 at 8:48
    
@Did: Done... ${}{}{}$ –  copper.hat Sep 6 '13 at 9:51
    
+1. Nice explanation. –  Did Sep 6 '13 at 10:12
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The matrix that governs the heptagonal isomorphs satisfies $A^3=I$.

 1   1  1        1   0  1            1  0  0
 0  0   1   ->   0  -1 -1   ->       0  1  0
 0 -1  -1        0   1  0            0  0  1

When this matrix is applied to numbers (i,a,b), where these are the chords of a unit-edge heptagon (1, 1.801937736, 2.2469796037), then the outcome comes to the chords of a unit edge heptagrams (7/3, 7/2), and back again.

But it is a matrix that satisfies this relation, and not in the form of cos(x), sin(x).

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Did you actually multiply them. @copper.hat –  wendy.krieger Sep 6 '13 at 21:41
    
Oops, I made a mistake when multiplying. Will delete comment. –  copper.hat Sep 6 '13 at 22:28
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