Algebraic or geometric multiplicity? I am reading a proof of the fact that

every linear transformation $L:V\to V$ can be represented by an upper triangular matrix $M$, with eigenvalues on the diagonal. And if the algebraic multiplicity of an eigenvalue $\lambda$ is $\alpha(\lambda)$, then $\lambda$ appears $\alpha(\lambda)$ times.

The proof is by induction on $m=dim(V)$. It's trivial when $m=1$. Now suppose it's true for all $1,2,3,\cdots,m-1$.
Then for $m$, because there exists at least one eigenvalue, let's call it $\lambda$. Then the author considers $M-\lambda I_V$, which is singular and has a range $W\subsetneq V$, with dimension $k<m$. Then he restricted the operation on $W$, which has dimension smaller than $m$ and can apply the induction assumption. He claimed that $V=W\oplus U$, where $U$ has dimension $l$ and $k+l=m$.
He claimed that $l$ is the algebraic multiplicity $\alpha(\lambda)$, but I think it should be the geometric multiplicity $\gamma(\lambda)$. My understanding is that $\gamma(\lambda)$ is the dimension of the subspace such that when acted on by $M-\lambda I_V$ gives the null vector, and so it should equal $l$.
What's wrong with my understanding? Thank you!
 A: You are correct in that $l$ is the geometric multiplicity of $\lambda$ and doesn't have to be the algebraic multiplicity of $\lambda$. Consider for example the matrix
$$ A = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda  \end{pmatrix} $$
and let $L(x) = Ax$ be the associated linear map on $\mathbb{F}^n$. Then we have $W = \left< e_1 \right>$ and so $k = 1$ and $l = 1$ but the algebraic multiplicity of $\lambda$ is $2$.
A: I think that the algebraic multiplicity of an eigenvalue $\lambda_0$ of a matrix $A$ is basically the multiplicity that $\lambda_0$ has in the characteristic polynomial of $A$. 
The geometric multiplicity of $\lambda_0$ should be the dimension of the eigenspace $E_{\lambda_0}(A)$, or the dimension of the nullspace $N(A-\lambda_0I)$. Hence I think that algebraic multiplicity is correct in the definition.
I think your doubt comes from the fact that, when a matrix is diagonalizable we know that the algebraic and the geometric multiplicity of an eigenvalue $\lambda_0$ are the same!
