Jordan decomposition of a 2 by 2 matrix

Question

I want to find the Jordan decomposition of the matrix $$ M= \begin{pmatrix} p & -\frac{p^2}{4} \\ 1 & 0 \end{pmatrix} $$ Wolfram Alpha gives me the decomposition $M = S \cdot J \cdot S^{-1}$, where $$ S= \begin{pmatrix} \frac{p}{2} & 1 \\ 1 & 0 \end{pmatrix}, \quad J= \begin{pmatrix} \frac{p}{2} & 1 \\ 0 & \frac{p}{2} \end{pmatrix}, \quad S^{-1}= \begin{pmatrix} 0 & 1 \\ 1 & -\frac{p}{2} \end{pmatrix}. $$ I have checked that $S^{-1}$ is indeed the inverse of $S$. However, if we multiply the three matrices we get $$ \begin{pmatrix} 0 & p \\ \frac{p}{2} & -\frac{p}{2} \end{pmatrix}. $$ Where is the problem? I suspect that $S$ is not correct. What would be the correct decomposition?

Answer 1

Looks like you made a mistake somewhere in the multiplication, $SJS^{-1}$ is actually equal to $\begin{pmatrix} p & -\frac{p^2}{4} \\ 1 & 0 \end{pmatrix}$.