Diagonalizability depending on a parameter

Eigenvalues. Since $A$ is upper triangular, the characteristic polynomial is $(1-\lambda)(2-\lambda)$, so the eigenvalues are $1$ and $2$.

Diagonalizability. The two eigenvalues are distinct, so their eigenspaces each have dimension $1$, and the sum is $2$. By the criterion, $A$ is diagonalizable for all $a \in \mathbb{R}$.

Explicit diagonalization. For $\lambda = 1$, solve $(A-I)v = 0$: $v_1 = (1,0)$. For $\lambda = 2$, solve $(A-2I)v = 0$: $v_2 = (a,1)$ (or $(a,1)$ if $a \neq 0$, $(0,1)$ if $a=0$). These two eigenvectors are linearly independent for all $a$, so $A = PDP^{-1}$ with
$$P = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}, \qquad D = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}.$$