Prove the matrix is not diagonalizable 
For any reals, $a, b$ show that 
$$D = \begin{bmatrix}
a & 1\\ 
0 &  b
\end{bmatrix}$$
  Is not diagonalizable.

I got that the eigenvalues are $\lambda_1 = a, \lambda_2 = b$
However, for $a \ne b$, there are $2$ distinct eigenvalues so $D$ must be diagonalizable right?
So what about $a= b$?
 A: In the case where $a = b$, we have only one eigenvalue $\lambda = a$. In order for the matrix to be diagonalisable, the algebraic multiplicity (let's denote it $\mu$) of the eigenvalue $\lambda$, where $\mu = 2$ in our example, must be equal to its geometric multiplicity $\gamma.$
We know by definition that $ \gamma = \dim \ker (D-\lambda I),$ with $$\ker (D-\lambda I) = \{ x \in \mathbb R^2: (D-\lambda I)x = 0\}.$$
However:
$$D-\lambda I = \begin{bmatrix} 0 & 1 \\ 0 & 0  \\ \end{bmatrix}.$$
We can easily notice that $\operatorname{rank} (D-\lambda I) = 1 \implies \dim \ker (D-\lambda I) = \gamma = 1$ (according to the rank-nullity theorem).
Thus, we have that $2 = \mu \neq \gamma = 1.$ Thus, whenever $a=b( = k)$, matrix $D$ is not diagonalisable, regardless the value of $k$.
A: $$
\mathbf{A} =
\left(
\begin{array}{cc}
 a & 1 \\
 0 & b \\
\end{array}
\right)
$$
Eigenvalues
$$
\lambda \left( \mathbf{A} \right) = (a, b)
$$
Eigenvectors:
$$
v_{1} =
\left(
\begin{array}{c}
 1 \\
 0
\end{array}
\right), \qquad
v_{2} =
\left(
\begin{array}{c}
 1 \\
 b-a
\end{array}
\right)
$$
(Good point by @Joonas Ilmavirta.)
