Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 following matrix: $$A = \left(\begin{array}{rrr} -1 & \hphantom{-}3 & \hphantom{-}0\\ 0 & 2 & 0\\ -3 & 3 & 2 \end{array}\right).$$ I've found that the eigenvalues are -1 and 2 (multiplicity 2). However, when I try to find the eigenvectors for the eigenvalue 2, I can only find one, as the augmented matrix $A-2I$ reduces to $$\left(\begin{array}{rrr} 1 & -1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right).$$ So is there any other way to diagonalise this matrix, or have I made a mistake somewhere?

share|cite|improve this question
Your matrix is diagonalizable. $A-2I = \left( \begin{align} -3 & 3 & 0\\\ 0 & 0 & 0\\\ -3 & 3 & 0\end{align} \right)$ – user17762 Feb 27 '11 at 20:12
You can check your work with the Online Matrix Calculator at – lhf Feb 27 '11 at 20:12
up vote 9 down vote accepted

First: not every matrix is diagonalizable; if the dimension of the eigenspace of a given eigenvalue is strictly smaller than the multiplicity of the eigenvalue, then the matrix is not diagonalizable (the condition is both necessary and sufficient for matrices whose characteristic polynomial splits: a matrix is diagonalizable if and only if the characteristic polynomial splits, and the geometric multiplicity of each eigenvalue equals the geometric multiplicity).

Second: this matrix is diagonalizable. You did everything right, but misinterpreted what $A-2I$ was telling you.

You are correct that your matrix has characteristic polynomial $\chi(t) = (2-t)^2(-1-t)$, so that $\lambda=2$ is an eigenvalue with multiplicity $2$, and $\lambda=-1$ is an eigenvalue with multiplicity $1$.

When you try to find the nullspace for $A-2I$ to find the eigenspace for $\lambda=2$, you get $$\left(\begin{array}{rrr} -3 & 3 & 0\\ 0 & 0 & 0\\ -3 & 3 & 0 \end{array}\right) \rightarrow \left(\begin{array}{rrr} 1 & -1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right),$$ exactly what you described, so far so good.

But now you've misinterpreted what you found: this matrix has rank $1$, so the nullspace has dimension two, exactly what you need! The matrix corresponds to the system that has the unique equation $x-y = 0$. This means that $z$ is completely free, and $x$ must equal $y$. So the solutions to this system are: $$\begin{array}{rcl} x & = &s\\ y & = &s\\ z & = &t \end{array}\qquad\quad\mbox{$s$ and $t$ arbitrary.}$$ So you get two linearly independent vectors: one corresponding to $s=0$ and $t=1$, and one corresponding to $s=1$ and $t=0$. So you can find two linearly independent eigenvectors corresponding to $2$: $(0,0,1)$ and $(1,1,0)$. Multiply them by $A$ to verify that they are indeed eigenvectors of $\lambda=2$.

share|cite|improve this answer

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.