(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.


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


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 $-1+2 \cos \theta_2$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.