Proof Checking: Decomposing a finite dimensional vector space as direct sum of mutually orthogonal invariant subspaces of an orthogonal endomorphism

I have been given the following question to do.

Suppose that α is an orthogonal endomorphism on the finite-dimensional real inner product space V . Prove that V can be decomposed into a direct sum of mutually orthogonal α-invariant subspaces of dimension 1 or 2.

I think I have a proof however I am not sure about it so can someone please check:

If $\alpha$ has a real eigenvector (with eigenvalue $\lambda$) then we can separate out the space spanned by that eigenvector and its orthogonal complement, then $\alpha$ is invariant on the eigenvector space and it restricts to the orthogonal complement of the eigenspace as well as if $u$ is in the orthogonal complements then $(\alpha(u),v) = (u,\alpha^{-1}(v)) = (u,\lambda^{-1}v) = \lambda^{-1}(u,v) = 0$ so we have reduced the problem to showing the result holds on the orthogonal complement which is a smaller space.

We can repeat this until we get a space on which $\alpha$ has no real eigenvectors left. Hence we only need to consider the case where $\alpha$ has no real eigenvectors (and so no real eigenvalues). Note that as $\alpha$ is real its characteristic polynomial's roots come in complex conjugate pairs so it can be factored as the product of quadratic polynomials. Say $\chi_{\alpha} = q_1q_2\cdots q_n$

Now note that there is a quadratic factor of $\chi_{\alpha}(\alpha)$ such that a vector in $V$ exists which is sent to zero by the factor. This is because $\chi_{\alpha}(\alpha)(v) = 0 \ \forall v \in V$ so there must be a factor $q_i$ which takes in a non-zero vector $w$ and gives out the zero vector. So for this vector we have $\alpha^2(w)+k_1\alpha(w)+k_2w = 0$ for some scalars $k_1,k_2$. So the space spanned by $w,\alpha(w)$ is invariant under $\alpha$ as $\alpha^2(w)$ can be written in terms of $w,\alpha(w)$. Similarly note that the inverse $\alpha^{-1}(w)$ can be written in terms of $w,\alpha(w)$ as $\alpha(w)+k_1w+k_2\alpha^{-1}(w) = 0$. The orthogonal complement of this space is also invariant under $\alpha$ as if $(u,v) = 0 \ \forall v \in \left< w,\alpha(w) \right>$, then $(\alpha(u),v) = (u,\alpha^{-1}(v)) = 0$ as $\alpha^{-1}(v) \in \left< w,\alpha(w) \right>$ (the space spanned by $w,\alpha(w)$) so $\alpha$ restricts to the orthogonal complement and we have a smaller space we need to show it for but by induction on the dimension (the base case $\dim(V) = 2$ is immediate from the definition) we get the result we need to prove.

If this is wrong can someone please point out the error I am making and potentially a give a hint or two?

• It's not clear what is meant by $\alpha^{-1}(v) \in \left< w,\alpha(w) \right>$. The left object is a vector, while the right is a scalar. – Christian Sykes Aug 6 '17 at 19:40
• @ChristianSykes Sorry, I meant that it is in the span of those two vectors and not an inner product, I have cleared it up. – Abdul Hadi Khan Aug 7 '17 at 7:13
• Ok, I understand your notation now. Your proof looks largely correct, but I think you should make some brief comment on what happens to the minimal polynomial of the induced endomorphism compared to the original. (Also, focusing on the minimal, rather the characteristic, polynomial probably is a better idea here in general.) To see why this matters, consider a linear transformation whose Jordan form has a Jordan block of size 3 or greater. Your argument would not work there. Why does it work here? – Christian Sykes Aug 7 '17 at 21:30
• @ChristianSykes I think it would not work as we wouldn't be guaranteed to find a vector which is sent to 0 by a quadratic factor of the characteristic polynomial (as for all we know everything could be mapped to zero by the cubic term in the characteristic polynomial), however we don't have the same problem with the minimal polynomial as all factors have to send something to zero or we could remove them and get a smaller minimal polynomial so any quadratic factor of the minimal polynomial is guaranteed to zero some vector. We also don't need Cayley Hamilton if we use the minimal polynomial. – Abdul Hadi Khan Aug 8 '17 at 10:10
• Also I think it works in this case as orthogonal matrices can't have a cubic factor in their minimal polynomial. This is because any orthogonal matrix is diagonalisable over $\mathbb{C}$ (special case of Hermitian matrix) so the minimal polynomial over $\mathbb{C}$ is a product of distinct linear factors (the eigenvalues) and as the minimal polynomial over $\mathbb{R}$ and $\mathbb{C}$ are the same we get that the minimal polynomial over $\mathbb{R}$ is a product of distinct linear and quadratic factors (by pairing off the complex linear factors with their conjugates). – Abdul Hadi Khan Aug 8 '17 at 10:26