Checking diagonalizability of a given $2\times 2$ matrix Let $A$ be the matrix $ A = \left(
                                 \begin{array}{cc}
                                  a  &  c\\
                                   0 & a \\
                                 \end{array}
                               \right)$ with $a, c \in \mathbb{R} $ . Can we impose any conditions on $ a $ and $c $ so that it may be diagonalized. In other words can we find matrix $P$ such that $PAP^{-1}$  is diagonal matrix.
i tried by taking certain conditions like if $a =0$ , $a = c$, and taking certain value of $a$ and $c$.
Then i came to conclusion that above matrix can not be diagonalized? Am i correct?
I want a proper explanation. 
 A: This matrix can be diagonalized if and only if $c=0$.  Note that the characteristic polynomial is $P(\lambda) = (\lambda-a)^2$, so $a$ is the only eigenvalue.  If $c \ne 0$ the null space of $A - a I = \pmatrix{0 & c\cr 0 & 0\cr}$ is only one-dimensional, being spanned by $\pmatrix{1 \cr 0\cr}$, so $A$ is not diagonalizable (a diagonalizable $n \times n$ matrix must have $n$ linearly independent eigenvectors).  If $c = 0$ the matrix is already diagonal.
A: Note that for the matrix you have, $$A = \begin{pmatrix} a & c \\ 0 & a\end{pmatrix},$$ the eigenvalue is $a$ and the algebraic multiplicity of the eigenvalue is $2$.
For the matrix to be diagonalizable, the geometric multiplicity of the eigenvalue must also be two i.e. the number of distinct eigenvectors corresponding to the eigenvalue $a$ must also be $2$. 
However, the eigenvalue $a$ yields only one eigenvector $x$ such that $$\begin{pmatrix} a & c \\ 0 & a\end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = a \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}.$$ Solving, we get the lone eigenvector as $$\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$
