Let $A \in SO(3)$ show that $1$ is always an eigenvalue. (Proof check and alternative proof?)

Question

Given $A \in SO(3)$ show that $1$ is always an eigenvalue. Where $SO(3)$ is the 3-dimensional special orthogonal group.

This is an exercise from Lie Groups and Algebras with Applications to Physics, Geometry, and Mechanics.

Intuitively it seems obvious. If I think of a rotating cilinder about some axis of rotation, then the eigenvector with eigenvalue $1$ would be a some vector on the axis of rotation.

The proof I found however, is more algebraic.

Can I get a proof check?

And is there a more intuistionictic proof? Or is the algebraic proof a result of the algebraic definition I used for $SO(3)$?

By definition of $SO(3)$ we have that the entries of $A$ are real, $A^T = A^{-1}$ and that $\text{det}(A) = 1$.

Lemma 1: the eigenvalues of $A$ are either $1$ or $-1$.

Proof: $$\langle v,v \rangle = \langle v,I v \rangle = \langle v,A^{-1}Av \rangle = \langle v,A^{T}Av \rangle = \langle {A^{T}}^T v,Av \rangle = \langle Av,Av \rangle =\langle \lambda v, \lambda v \rangle = \lambda^2 \langle v, v \rangle $$ So $\lambda^2 = 1$, so $\lambda$ is either $1$ or $-1$. QED

Lemma 2: $\text{det}(A) = \lambda_1 \lambda_2 \lambda_3$.

Proof: Let $v$ be an eigenvector with eigenvalue $\lambda$, then $$Av = \lambda v$$ $$Av = \lambda I v$$ $$(A - \lambda I) v = 0.$$ Since, $v$ is non-zero, the matrix $A - \lambda I$ is non-invertible, which yields the characteristic equation: $$\text{det}(A-\lambda I) = 0.$$ Since we already know that all eigenvalues are real, the characteristic equation splits in $R$: $$\text{det}(A-\lambda I) = (\lambda_1 - \lambda)(\lambda_2 - \lambda)(\lambda_3 - \lambda).$$ Setting $\lambda = 0$ gives: $$\text{det}(A) = \lambda_1 \lambda_2 \lambda_3.$$ QED

For a contradiction, assume that all eigenvalues equal $-1$, then by lemma 2 we get $\text{det}(A) = -1$, which is a contradiction. So, at least one eigenvalue is not equal to $-1$. By lemma 1, this eigenvalue must be equal to 1. QED.

Answer 1

As the comments said, your Lemma 1 is wrong. However, your proof can be correct after slightly modified.

The correct statement of Lemma 1 is:

Lemma 1: the REAL eigenvalues of $A$ are either 1 or −1.

and this can be proved using your method.

You have solved the problem when the characteristic equation $\mathrm{det}(A−λI)=0$ have 3 real roots. Otherwise, if the characteristic equation have one real root $a$ and a pair of complex roots $b\pm ci$, we can get $$1 = \mathrm{det}(A) = a(b+ci)(b-ci) = a(b^2+c^2).$$ Thus $a=\frac{1}{b^2+c^2}$ is positive and can only be 1.