Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
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 $$\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.