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I'd like to find the angle of rotation of the following matrix $A= \begin{bmatrix} -\frac{1}{3} & \ast & \ast \\ \ast & -\frac{1}{3} & \ast \\ \ast & \ast & -\frac{1}{3} \end{bmatrix}$

I know that because it is an orthogonal matrix there exists some $T \in GL(3, \mathbb{C}$ so that $T^{-1}AT = \begin{bmatrix} \pm 1 & 0 &0 \\ 0 & cos(\theta) & -sin(\theta) \\ 0 & sin(\theta) & cos(\theta) \end{bmatrix} $ where $\theta$ is the angle of rotation. The problem is that I would've done this by finding the eigenvalues and now I'm unsure as to how else I can approach this.

If anyone could give me a hint as to what I need to look at that'd be greatly appreciated!

Widawensen
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Longeyes
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1 Answers1

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A rotation matrix $R$ in $SO_3(\Bbb{R})$ is conjugate to a matrix of the form $$ A=\left(\begin{array}{crr} 1&0&0\\ 0&\cos\theta&-\sin\theta\\ 0&\sin\theta&\cos\theta \end{array}\right). $$ In other words there is a matrix $T$ such that $R=TAT^{-1}$.

Hint: Combine the following two facts (prove these first, if you haven't seen them before)

  • $\operatorname{tr}(A)=1+2\cos\theta$,
  • $\operatorname{tr}(TAT^{-1})=\operatorname{tr}(A).$

Recall that the trace of a square matrix $A$, $\operatorname{tr}(A)$, is the sum of its diagonal entries (and also the sum of its eigenvalues with correct multiplicities).

I guess that for the sake of completeness it behooves me to also show that rotation matrices with the prescribed diagonal actually exist. To that end consider $$ R=\frac13\left(\begin{array}{ccc} -1&2&2\\ 2&-1&2\\ 2&2&-1 \end{array}\right). $$ It is easy to check that the rows (or columns) of $R$ form an orthonormal system. It is also easy to check that $\det R=1$, so the system is right-handed.

Jyrki Lahtonen
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