Diagonalizability depending on a parameter

Eigenvalue. The matrix is upper triangular with both diagonal entries $1$, so $\lambda=1$ is the only eigenvalue, with algebraic multiplicity $2$.

Eigenspace.
$$A_a-I = \begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}.$$
- If $a\neq0$: the kernel is $\{(x,0)\}$, dimension $1$.
- If $a=0$: $A_0=I$, and the kernel is $\mathbb{R}^2$, dimension $2$.

Conclusion.
- If $a\neq0$: geometric multiplicity $1<2$, so $A_a$ is not diagonalizable.
- If $a=0$: $A_0=I$ is already diagonal, hence diagonalizable.