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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.

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  • $\begingroup$ You are almost there. Use the fact that $\overline{\lambda}$ is an eigenvalue of $T$ whenever $\lambda$ is an eigenvalue. $\endgroup$ – Giuseppe Negro May 21 '12 at 18:32
  • $\begingroup$ If the eigenvalue $1$ has multiplicity $>1$, then it must, in fact, have multiplicity $3$. $\endgroup$ – copper.hat May 21 '12 at 19:15
  • $\begingroup$ 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 ? $\endgroup$ – Ester May 22 '12 at 2:31
  • $\begingroup$ @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. $\endgroup$ – Phira May 22 '12 at 11:28
  • $\begingroup$ How do you prove that every orthogonal matrix is diagonalisable ? I knew that every unitary matrix is diagonalisable . $\endgroup$ – Ester May 23 '12 at 3:26
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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 $T$.

Q.E.D

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  • $\begingroup$ @ 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 ? $\endgroup$ – Ester May 23 '12 at 13:44
  • $\begingroup$ @ 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. $\endgroup$ – users31526 May 23 '12 at 14:14

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