I'm trying to apply an algorithm I found online to compute the Jordan form of the following matrix
$A = \begin{bmatrix}0&-1\\1&0\end{bmatrix}$
- Find characteristic polynomial and eigenvalues of A
After some calculation...
$\lambda = +_- i$ and
- Find null basis of A
Well the null basis of A are vectors that when A operates on them, the whole thing goes to zero..
$A\vec v =0$
Hmm! Why do I only get $\vec v$ = 0 as the solution?
- Find $J = B^{-1} A B$, that's your Jordan matrix
Can someone point out where in this algorithm I need to adjust to find the Jordan matrix?
To find the eigenvectors, we have $[A - \lambda_i I]v_i = 0$.
The eigenvalues are:
$$\lambda_{1,2} = \pm i$$
So, for $\lambda_1$, we have $[A - i I]v_1 = 0$, and a row-reduced-echelon-form leads to:
$$\begin{bmatrix} 1 & -i \\ 0 & 0 \\ \end{bmatrix}v_1 = 0$$
This leads to:
$$v_1 = \begin{bmatrix} i \\ 1 \end{bmatrix}$$
Since these are complex eigenvalues, the eigenvectors are complex conjugates, hence:
$$v_2 = \begin{bmatrix} -i \\ 1 \end{bmatrix}$$
The eigenvalues are unique, so we knew the Jordan Normal Form, $J$, from that alone (it is a diagonal matrix made up of the unique eigenvalues), but we can also write:
$$J = P^{-1}AP$$
$P$ is made up of the eigenvectors for each $\lambda_i$.
We have:
$$J = \begin{bmatrix} -i & 0 \\ 0 & i \\ \end{bmatrix}, P = \begin{bmatrix} -i & i \\ 1 & 1 \\ \end{bmatrix}$$
Notice that the eigenvectors in $P$ track the eigenvalues in $J$. Also, you could have written the eigenvalues in any order you like, so long as the eigenvectors track.