0
$\begingroup$

I have a matrix $$B=\begin{pmatrix}1&2&3\\0&4&5\\0&0&6\end{pmatrix}$$

I have calculated the eigenvectors: $$\{\begin{pmatrix}-1\\0\\0\end{pmatrix},\begin{pmatrix}-\frac{2}{3}\\-1\\0\end{pmatrix},\begin{pmatrix}-\frac{8}{5}\\-\frac{5}{2}\\-1\end{pmatrix}\}$$

However, I am trying to find the matrix $S$ such that $$SBS^{-1}=\begin{pmatrix}1&0&0\\0&4&0\\0&0&6\end{pmatrix}$$

(The jordan normal form)

I have been told that the columns of $S$ are the eigenvectors of $B$, but $$\left( \begin{array}{ccc} -1 & -\frac{2}{3} & -\frac{8}{5} \\ 0 & -1 & -\frac{5}{2} \\ 0 & 0 & -1 \\ \end{array} \right)\left( \begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{array} \right)\left( \begin{array}{ccc} -1 & \frac{2}{3} & -\frac{1}{15} \\ 0 & -1 & \frac{5}{2} \\ 0 & 0 & -1 \\ \end{array} \right)=\left( \begin{array}{ccc} 1 & 4 & \frac{13}{3} \\ 0 & 4 & 10 \\ 0 & 0 & 6 \\ \end{array} \right)$$

Do I need scalar multiples of the eigenvectors?

$\endgroup$
3
$\begingroup$

Using the eigenvectors gives a matrix such that $$B=SJS^{-1}\ ,$$ in other words $$J=S^{-1}BS\ .$$ So you have used $S$ where you should have used $S^{-1}$, and vice versa.

$\endgroup$
  • $\begingroup$ Oh lord! You just saved me a pile of time. Thank you so much! $\endgroup$ – Dia McThrees Nov 12 '14 at 2:53
  • 1
    $\begingroup$ If you think carefully it is obvious that it must be this way round. Because finding $BS$ involves multiplying $B$ by columns of $S$, in other words $B$ times eigenvectors, which is the whole point of eigenvectors. But $SB$ means nothing in particular. $\endgroup$ – David Nov 12 '14 at 2:55
  • $\begingroup$ Would multiples of the eigenvectors still result in the Jordan form? $\endgroup$ – Dia McThrees Nov 12 '14 at 2:56
  • 1
    $\begingroup$ A (non-zero) multiple of an eigenvector is still an eigenvector, so yes. In the case where the same eigenvalue occurs more than once, you have to be a bit careful. $\endgroup$ – David Nov 12 '14 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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