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Let $T:\mathbb{R^3} \rightarrow \mathbb{R^3}$ be an orthogonal transformation such that $\det T = 1$ and $T\neq I$. Let S be the unit sphere in $\mathbb{R^3}$. I need to show that $T$ fixes exactly two points on S.
What I think is, if I can show $T$ has eigenvalue $1$ with geometric multiplicity $1$, then I am done.The possible eigenvalues of $T$ are:
1)$1,-1,-1$
2)$1,a+ib,a-ib$ where $a,b\in\mathbb R$ and $a^2+b^2=1$
3) all the eigenvalues are $1$.
Now I am struggling with the case 3. How can I show that the case 3 is not possible? I think I have to use the fact $T\neq I$, but don't know how.

José Carlos Santos
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simu tiyam
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  • Any real orthogonal matrix is normal so by spectral theorem all eigenvalues (that exist) are semi-simple -- i.e. geometric multiplicity equal algebraic multiplicity – user8675309 Apr 30 '21 at 17:40
  • But that's for complex orthogonal matrix,right? – simu tiyam Apr 30 '21 at 18:07
  • You need to be a lot more careful with definitions -- the answer is no. The matrices $\in O_n(\mathbb C)$ are not normal in general. On the other hand $U_n(\mathbb C)$ only has normal matrices in it --i.e. unitary matrices are normal. And real orthogonal matrices may be viewed as a special case of unitary. – user8675309 Apr 30 '21 at 19:34
  • I know when we consider T as a complex matrix, then by spectral theorem it’s diagonalizable, i.e, algebraic and geometric multiplicity is same for any eigenvalue of T. But how are you saying the same thing for real orthogonal matrix? Please explain – simu tiyam May 01 '21 at 01:35
  • 3x3 orthogonal matrix T has real eigenvalue 1 with algebraic multiplicity 3. When we consider T as complex matrix, we know it’s diagonalizable by spectral theorem . That's geometric multiplicity is also 3. Suppose the independent eigenvectors { (1,0,0),(0,i,0),(0,0,i)}. But how are you saying that T as real matrix will also have geometric multiplicity 3? – simu tiyam May 01 '21 at 01:55
  • Something fundamental seems to be missing. Here's one approach: $B:=T-I$ with your real orthogonal $3\times 3$ matrix $T$ that has eigenvalue 1 with algebraic multiplicity 3. Now mimicking https://math.stackexchange.com/questions/3777401/does-real-dimension-equal-rational-dimension/ . Your ground field is $\mathbb R$ and the extension is $\mathbb C$. But $\text{rank}\big(B\big) = 0$ over $\mathbb C$ and is the same over $\mathbb R$ (per link). Now apply rank nullity, working over $\mathbb R$: $\dim \ker B = 3=$ geometric multiplicity of eig 1 for $T$. – user8675309 May 01 '21 at 17:21

1 Answers1

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If all eigenvalues are $1$, then there are three linealy independent vectors $v_1$, $v_2$ and $v_3$ such that$$T(v_i)=v_i\text{ for each }i\in\{1,2,3\}.\tag1$$But then $\{v_1,v_2,v_3\}$ is a basis of $\Bbb R^3$ and it follows from $(1)$ that $T$ is the identity map.

José Carlos Santos
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  • Can't geometric multiplicity be less than 3? – simu tiyam Apr 30 '21 at 16:48
  • For $3\times3$ matrices in general, yes; for orthogonal matrices, no. – José Carlos Santos Apr 30 '21 at 16:49
  • I was not aware of that. Can you tell me why is that? Or give me a hint maybe. – simu tiyam Apr 30 '21 at 16:53
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    Suppose that $T(v)=v$, for some $v\ne0$. Consider the set $v\perp={w\in\Bbb R^3|mid\langle v,w\rangle=0}$. Then (since $T$ is orthogonal), $T\left(v^\perp\right)\subset v^\perp$. But $\dim v^\perp=2$ and $T|{v^\perp}$ is orthogonal and it has determiant $1$. Therefore, it is a rotation, with respect to some angle $\theta\in[0,2\pi)$. If $\theta=0$, then $T=\operatorname{Id}$. Otherwise, $1$ is not an eigenvalue of $T|{v^\perp}$, and therefore the geometric multiplicity of $1$ is $1$. – José Carlos Santos Apr 30 '21 at 16:57
  • I don’t know much about rotation. It’s little hard for me to understand. So for any odd orthogonal matrix algebraic and geometric multiplicity regarding an eigenvalue will be same,right? I will look into that. – simu tiyam Apr 30 '21 at 17:18
  • Yes, that is correct. – José Carlos Santos Apr 30 '21 at 17:21