TITLE: Examples for proof of geometric vs. algebraic multiplicity
QUESTION [12 upvotes]: Here you see a supposedly easy proof of a well-known theorem in linear algebra:

Although I know I should understand this, I don't :-(
Obviously there are too many indices and stuff, so I don't see the forest for the trees.
Please could anyone give me some examples which show what is going on in this proof? I really need a step-by-step tour which connects this proof with some examples, I guess...
(You could also give references which provide this kind of step-by-step approach, unfortunately most references I found were even more abstract...)
Thank you in advance!

REPLY [2 votes]: Basically, the idea is in the previous answers. Nevertheless, I think the following proof is easier to understand. Let's prove the following:
Fact: Suppose $\lambda_0$ is an eigenvalue of $A$ and with geometric multiplicity $k$, then its algebraic multiplicity is at least $k$.
Proof: By the assumption, we can find an orthonormal basis for the subspace $\mathrm{ker}(A-\lambda_0 I)$. Let us denote them by $u_1,u_2,\cdots,u_k$. We can then find an orthonormal basis for $\mathbb{R}^n$ by completing this basis. Denote them by $\{u_1,u_2,\cdots,u_k,u_{k+1},\cdots,u_n\}$. Denote the matrix formed by these vectors (as columns) by $S$. Then let $\lambda$ be a variable, we easily verify
$$
(A-\lambda I)S = \pmatrix{(\lambda_0-\lambda)u_1, \cdots (\lambda_0-\lambda)u_k, v_{k+1}(\lambda),\cdots, v_n(\lambda)}.
$$
The right hand side above is a matrix whose first $k$ columns are of the form
$$
(\lambda_0-\lambda) u_j, j=1,\cdots,k.
$$
The remaining columns have entries depending on $\lambda$ (in fact the entries are affine functions of $\lambda$) but we don't care their exact form.
Take the determinant on the first equation in the proof. We get
$$
\det(A-\lambda I) \cdot \det(S) = (\lambda_0-\lambda)^k \det\pmatrix{u_1,\cdots,u_k,v_{k+1}(\lambda),\cdots,v_n(\lambda)} = (\lambda_0-\lambda)^k g(\lambda)
$$
where $g(\lambda)$ is a polynomial in $\lambda$.
Since $\det(S) = \pm 1$, we see clearly that $\lambda_0$ is a root of the characteristic polynomial and its algebraic multiplicity is at least k. QED