Examples for proof of geometric vs. algebraic multiplicity 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!
 A: If the geometric multiplicity of $\lambda\in\mathbb C$ for the $n$-by-$n$ matrix $A\in M_n(\mathbb C)$ is at least $k$, then there are $k$ linearly independent eigenvectors $v_1$, $v_2$, $\dots$, $v_k$ corresponding to the eigenvalue $\lambda$. In other words, the system of equations $$(\lambda I-A)x=0$$ has a solution space of dimension at least $k$. This means that the rank of $\lambda I-A$ is at most $n-k$. It follows that by doing row operations (both permutating rows and adding to a row a multiple of another row) you can transform $\lambda I-A$ into a matrix $B$ whose last $k$ rows are zero. 
Now let $t$ be a variable (as opposed to a scalar, like $\lambda$) and perform exactly the same row operations on the matrix $t I-A$, whose entries are now polynomials in $t$. This gives you a matrix $C$, with entries also polynomials in $t$, which becomes $B$ when you evaluate these polynomials at $\lambda$. In particular, all the matrix entries of $C$ in the last $k$ rows are polynomials which vanish at $\lambda$, so they are divisible by $t-\lambda$. It is immediate then that the determinant of $C$ is divisible by $(t-\lambda)^k$. 
Indeed, you can pull the common factor $t-\lambda$ from each of those rows outside of the determinant.
Since the determinants of $t I-A$ and $C$ at worst differ by a constant multiple, this means that the characteristic polynomial of $A$, namely $\det(t I-A)$, is divisible by $(t-\lambda)^k$.
Notice that this is not the same argument as in the image you posted. I happen to like it much more, though :)
 
An example: Let $$A=\left(
\begin{array}{cccc}
 -1 & 2 & 0 & 1 \\
 -3 & 4 & 0 & 1 \\
 -\frac{2}{3} & 0 & 2 & \frac{2}{3} \\
 -3 & 2 & 0 & 3
\end{array}
\right).$$ If on the matrix $2 I-A$ I substract the first row from the second and the fourth, then substract $2/9$ times the first row from the third, and finally interchange the second and third rows, I get the matrix $$\left(
\begin{array}{cccc}
 -3 & 2 & 0 & 1 \\
 0 & -\frac{4}{9} & 0 & \frac{4}{9} \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0
\end{array}
\right).$$ This is a matrix of rank $2$ ---and the last two rows zero--- so in particular the geometric multiplicity of $2$ as an eigenvalue of $A$ is $2$.
If I now repeat the same operations on the matrix $tI-A$, I get $$\left(
\begin{array}{cccc}
 -t-1 & 2 & 0 & 1 \\
 -\frac{2}{9} (-t-1)-\frac{2}{3} & -\frac{4}{9} & 2-t &
   \frac{4}{9} \\
 t-2 & 2-t & 0 & 0 \\
 t-2 & 0 & 0 & 2-t
\end{array}
\right).$$ As predicted above, this matrix has in its last two rows polynomials which vanish at $2$. Computing the determinant $$\left|
\begin{array}{cccc}
 -t-1 & 2 & 0 & 1 \\
 -\frac{2}{9} (-t-1)-\frac{2}{3} & -\frac{4}{9} & 2-t &
   \frac{4}{9} \\
 t-2 & 2-t & 0 & 0 \\
 t-2 & 0 & 0 & 2-t
\end{array}
\right|$$ is very easy, and you can actually see without computing it in full that it is a polynomial divisible by $(t-2)^2$. This means that the algebraic multiplicity of $2$ as an eigenvalue is at least $2$.
A: 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
