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(We're in $\mathbb{R}^3$)

What can we say about type of linear isometry $F : \mathbb{R}^3 \to \mathbb{R}^3$ if trace of $\mathrm{m} (F)$ is $-2$ or $\frac{1}{\sqrt{2}}$ or $\sqrt{2}$? Which one of these three cases says anything about type of linear isometry (is it rotation, symmetry or something else), and which one doesn't tell anything?

So far I just figured out, that any 3x3 matrix has at least one real eigenvalue $|\lambda| = 1$ and of course $\mathrm{tr} (\mathrm{m} (F)) = \mathrm{tr} (PJP^{-1}) = \mathrm{tr} (J)$, where $J$ is either $\begin{pmatrix} \lambda & 0 & 0 \\ 0 & \mu & 0 \\ 0 & 0 & \eta \end{pmatrix}$ or $\begin{pmatrix} \lambda & 0 & 0 \\ 0 & \mu & 1 \\ 0 & 0 & \mu \end{pmatrix}$ or $\begin{pmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{pmatrix}$. Thus in case of trace $-2$ is either

  • $3\lambda = -2$
  • or $\lambda + 2\mu = -2$
  • or $\lambda + \mu + \eta = -2$

First case we rule out, cause only eigenvalue has to be $1$ or $-1$, so that can't hold. Second case turns into

  • $\lambda = 1$ and $\mu = -\frac{1}{2}$ or $\lambda = -1$ and $\mu = -\frac{3}{2}$
  • $\lambda = 0$ and $\mu = -1$ or $\lambda = -4$ and $\mu = 1$

And last case is just $\lambda + \mu = -1$ or $\lambda + \mu = -3$

I don't know what can I do with that information, how to determine type of linear isometry using that. Any hints would be much appriciated.

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It is not necessary at all to use Jordan form.

You only need two results :

1) The eigenvalues of an isometry have a unit modulus.

2) In $\mathbb{R^3}$, the characteristic equation of the corresponding matrix has degree 3, thus with

  • either 3 real values, which must then be either $1$ (order of multiplicity n) and/or $-1$ (order of multiplicity $(3-n)$), thus with trace

$$n1+(3-n)(-1) \ = \ 2n-3 ;$$

Otherwise said, in this case, the trace can take only 4 values

$$\{-3,-1,1,3\},$$

none of them corresponding to the desired traces :

$$t_1=-2, t_2=\dfrac{1}{\sqrt{2}}, t_3=\sqrt{2}. \tag{1}$$

  • or one real value ($+1$ or $-1$) and two complex conjugate ones $e^{\pm i \theta}$ giving a trace equal either to $1+2 \cos \theta$ (by using Euler formula $e^{i \theta}=\cos \theta+i \sin \theta$), covering the range $(-1,3)$, or $-1+2 \cos \theta$ covering the range $(-3,1)$ thus giving solutions for all the values given in (1).

Let us consider the case of $t_1=-2$ : one sees that the eigenvalues are necessarily $\{-1, e^{i \theta}, e^{-i \theta}\}$ (with $\theta=2\pi/3$), the corresponding matrix being :

$$\begin{pmatrix}-1&0&0\\0&\cos(\theta)&-\sin(\theta)\\0&\sin(\theta)&\cos(\theta) \end{pmatrix}=\begin{pmatrix}-1&0&0\\0&1&0\\0&0&1 \end{pmatrix}\begin{pmatrix}1&0&0\\0&\cos(\theta)&-\sin(\theta)\\0&\sin(\theta)&\cos(\theta) \end{pmatrix}$$

which can be characterized as a rotation in plane $yOz$ followed by the orthogonal symmetry with respect to this plane (in either order, in fact).

I leave you the two other cases.

Remark dealing with the subrange $(-1,1) \subset (-3,3)$ : if the trace is in this reduced range, one can express it in two different ways $1+2 \cos \theta_1$, or $-1+2 \cos \theta_2$.

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There are basically two kinds of linear isometry of $\mathbb R^3$: rotations (including the identity transformation which is a rotation by angle $0$) and improper rotations (including reflection across a plane which is an improper rotation by angle $0$, and reflection through the origin which is an improper rotation by angle $\pi$). A rotation of $\mathbb R^3$ by angle $\theta$ has eigenvalues $1$, $e^{i\theta}$ and $e^{-i\theta}$, so its trace is $1 + 2 \cos(\theta) \in [-1,3]$. An improper rotation by angle $\theta$ has eigenvalues $-1$, $e^{i\theta}$ and $e^{-i\theta}$ so its trace is $-1 + 2 \cos(\theta) \in [-3,1]$. So if the trace is $-2$ it can only be an improper rotation with $\cos(\theta) = -1/2$, if the trace is $\sqrt{2}$ it can only be a rotation with $\cos(\theta) = (\sqrt{2}-1)/2$, and if the trace is $1/\sqrt{2}$ it could be either.

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