# Find matrix that implements Jordan normal form

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?

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.
• 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. – David Nov 12 '14 at 2:55