Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $T\colon\mathbb{R}^3\to\mathbb{R}^3$ be an orthogonal linear transformation so that $\det T = 1$ and $T$ is not the identity transformation. Consider $S = \{ ( x , y , z ) : x^2 + y^2 + z^2 = 1 \}$. Show that $T$ fixes exactly $2$ points on $S$.

I have proceeded like this. $T\colon \mathbb{R}^3\to\mathbb{R}^3$ implies $T$ has at least one real eigenvalue (as the characteristic polynomial is of degree $3$). We know this real eigenvalue must be $1$ or $-1$. Then I am trying to show it has the eigenvalue $1$ of multiplicity $1$. Thanks for any help.

share|cite|improve this question
Dear Sopu, Kindly refer here (…) on how to typeset (i.e. how to write equations etc) on this website. This is done so that the equations and other math appear in a nice way. – user17762 May 21 '12 at 17:54
Is this homework? If so, it should be tagged as such... – Shai Deshe May 21 '12 at 17:59
No , this is not homework . – Ester May 21 '12 at 18:01
You are almost there. Use the fact that $\overline{\lambda}$ is an eigenvalue of $T$ whenever $\lambda$ is an eigenvalue. – Giuseppe Negro May 21 '12 at 18:32
If the eigenvalue $1$ has multiplicity $>1$, then it must, in fact, have multiplicity $3$. – copper.hat May 21 '12 at 19:15
up vote 1 down vote accepted

Since $T:\mathbb R^{3}\rightarrow \mathbb R^{3}$ is an orthogonal linear transformation (over the underline field $\mathbb R$) hence all its eigen values are of modulus 1 (exercise: Hint use $(T(x),T(x))$ and definition of orthogonality).

Now $det(T)=1$ and $T$ is non identity gives you for each eigen value and respective eigen space consists at least one non-zero eigen vector which proves multiplicity of each eigen value has at least one.

Since you have chosen $\mathbb R^{3}$ so characteristic polynomial has degree 3. So the polynomial of $T$ must looks like $a(x-c)(x-z)(x-\overline z)$ where $a\in \mathbb R, c=\pm 1$ (but $c\neq -1$ as a=1 and $det(T)=c.z.\overline z$) and $z,\overline z\in \mathbb C$ and $z\neq \overline z$. Since $z$ has multiplicity at least $1$ and $\overline z$ has multiplicity at least $1$ hence $c=1$ has multiplicity exactly (at most and at least ) $1$.

Hence eigen space of $c$ has exactly one non-zero vector say $x$ then $\frac {x}{||x||}, -\frac {x}{||x||}$ are only $2$ points $\in S$ which are fixed by $T$.


share|cite|improve this answer
@ Kuashik How are you predicting this form of the polynomial ? It can be that all the eigenvalues are 1 , one of them 1 and the rest -1 . If all the eigenvalues are 1 , then how can you conclude from your result that the algebraic muktiplicity is 1 ? – Ester May 23 '12 at 13:44
@ Sopu: In my answer multiplicity means geometric multiplicity. Your case will not disturb by the argument in para 2. But I am wrong in the part "$z\neq \overline z$" in para 3. – users31526 May 23 '12 at 14:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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