Compute the Jordan normal form of a $2 \times 2$ matrix

Question

For $t \in \Bbb R$, consider

$$A = \begin{pmatrix} 0 & t \\ 0 & 0 \\ \end{pmatrix}$$

Compute the Jordan normal form of $A$. Distinguish between $t=0$ and $t \neq 0$.

What I have done so far. I tried to find the eigenvalues when $t=0$ or any other number and I got $2$ eigenvalues which both are $0$. How do I proceed from here? Thanks.

Answer 1

If $t=0$, then $A=0$ the zero matrix which is already in Jordan form.

If $t\neq 0$, then define $P=\begin{pmatrix} \frac{1}{t}& 0 \\ 0& 1\end{pmatrix}$. Note that $P^{-1}=\begin{pmatrix} t& 0 \\ 0& 1\end{pmatrix}$. Direct calculation reveals $PAP^{-1}=\begin{pmatrix} 0& 1\\ 0& 0\end{pmatrix}$, which is in Jordan form.

Answer 2

Hint:

• if $t=0$ then $A$ is already in a Jordan normal form.
• if $t \neq 0$ then by considering the basis $((t,0),(0,1))$ you can show that: $$A \sim \begin{pmatrix}0&1\\0&0 \end{pmatrix}$$