I'm trying to apply an algorithm I found online to compute the Jordan form of the following matrix

$A = \begin{bmatrix}0&-1\\1&0\end{bmatrix}$

  1. Find characteristic polynomial and eigenvalues of A

After some calculation...

$\lambda = +_- i$ and

  1. 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?

  1. 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?

  • $\begingroup$ For $2$, note that $\det(A)\ne 0$. $\endgroup$ – Adam Hughes Mar 6 '15 at 2:04

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.


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