Can an eigenvalue (of an $n$ by $n$ matrix A) with algebraic multiplicity $n$ have an eigenspace with fewer than $n$ dimensions? Is it possible for a matrix with characteristic polynomial $(λ−a)^3$ to have an eigenline (one-dimensional eigenspace)?
I know that geometric multiplicity can generally be smaller than algebraic multiplicity. But I was wondering if algebraic multiplicity $n$ might be a special case. This question is motivated by my earlier one The greatest possible geometric multiplicity of an eigenvalue, where I learnt that $A$ has an $n$-dimensional eigenspace iff. $A=\lambda I$.
 A: Sure. Consider the matrix 
$$\begin{pmatrix} 1 & 1 \\
0 & 1\end{pmatrix}$$
which has a single eigenvalue $1$ of multiplicity $2$, yet the only solutions to the equation
$$\begin{pmatrix} 1 & 1 \\
0 & 1\end{pmatrix}\begin{pmatrix} x \\
y\end{pmatrix}=\begin{pmatrix} x +y \\
y\end{pmatrix}=\begin{pmatrix} x \\
y\end{pmatrix}$$
come when $y=0$, and so the eigenspace is $1$-dimensional.
A: Yes. Take for example
$$A = \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$$
whose characteristic polynomial is $(\lambda-1)^3$, but whose eigenspace is one-dimensional (spanned by $(1,0,0)$.)
If the matrix is symmetric, this can't happen though. (This is a nice exercise to show.)
A: Algebraic multiplicity $n$ is not a special case. Take the following operator for example:
$$
T(w, z) = (z, 0)
$$
It has the following matrix in the standard base:
$$
\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}
$$
The characteristic polynomial is $z^2$. Yet, $\operatorname{null}T$ is one-dimensional (spanned by $(1, 0)$).
Algebraic multiplicity of an eigenvalue $\lambda$ is equal to $\operatorname{null}(T - \lambda I)^{\operatorname{dim}V}$. In this example, $\operatorname{null}T^2$ does indeed have 2 dimensions.
