# Non-identity Orthogonal Transformation

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.

• You are almost there. Use the fact that $\overline{\lambda}$ is an eigenvalue of $T$ whenever $\lambda$ is an eigenvalue. Commented May 21, 2012 at 18:32
• If the eigenvalue $1$ has multiplicity $>1$, then it must, in fact, have multiplicity $3$. Commented May 21, 2012 at 19:15
• Yes , I have done that , but I cannot prove that if all the eigenvalues of T be 1 then T is the identity transfirmation . Can you please help me ? Commented May 22, 2012 at 2:31
• @Sopu An orthogonal matrix is diagonalizable. If all eigenvalues are 1, it is diagnoalizable to the identity matrix, but this implies that it is the identity matrix. Commented May 22, 2012 at 11:28
• How do you prove that every orthogonal matrix is diagonalisable ? I knew that every unitary matrix is diagonalisable . Commented May 23, 2012 at 3:26

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 $T$.
• @ 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. Commented May 23, 2012 at 14:14