8

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?

Szekely
  • 81
  • 2

3 Answers3

2

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.

W. Cadegan-Schlieper
  • 742
2

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.

Did
  • 279,727
copper.hat
  • 172,524
1

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).

wendy.krieger
  • 6,856