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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.

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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.

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Now it is clear to me sir –  srijan May 9 '12 at 7:31

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 $$\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

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i got the point. thanks –  srijan May 9 '12 at 7:31
    
Note that $$\mathbf A=\frac1{c}\begin{pmatrix}a/c&1\\&a/c\end{pmatrix}$$, showing that $\mathbf A$ is a scalar multiple of a Jordan block, the canonical example of a nondiagonalizable matrix... –  J. M. May 9 '12 at 8:45

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