Proving Algebraic Multiplicity doesn't equal Geometric Multiplicity

Question

I'm trying to show that the upper triangular matrix below is not diagonalisable.

Note, each $\star$ can represent any complex number.

$A = \begin{bmatrix} 1 & 100 & \star & \dots & \star \\ 0 & 1 & \star & \dots & \star\\ 0 & 0 & 201 & \dots & \star \\ \vdots & \vdots & \vdots & \ddots & \star\\ 0 & 0 & 0 & \dots & \star \\ \end{bmatrix}$

I know that the algebraic multiplicity of $1$ is at least $2$ and hope to prove that the geometric multiplicity of $1$ is less than or equal to $1$.

I've tried using rank-nullity theorem to prove this, but have had no success.

Any guidance would be greatly appreciated.

Answer 1

Let $e_1, \dotsc, e_n$ be the standard basis and $U=\langle e_1, e_2 \rangle$.

If $A$ were diagonalisable, then so were $A_{|U}=\begin{pmatrix}1&100\\0&1\end{pmatrix}$.