# Archive for June, 2013

1. First thing we do is find out the spectrum of $$A$$, that is, its eigenvalues. We shall mark them as $$\text{sp}A=\{\lambda_1, \lambda_2, \lambda_3, \cdots, \lambda_n\}$$. We do this however we can, usually by finding the zeros of the characteristic polynomial, that is solving the equation $$\det(A-\lambda I) = 0$$ for $$\lambda$$
2. In the next step we will be examining the sequence of generalized eigenspaces (i.e. $$\ker(A-\lambda I)^n$$ for a given $$n$$) for each $$\lambda \in \text{sp}A$$. $$\begin{gather} \ker(A-\lambda I) \\ \ker(A-\lambda I)^2 \\ \ker(A-\lambda I)^3 \\ \vdots \end{gather}$$ We calculate each kernel by simply solving the equation $$(A-\lambda I)^n \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n \end{bmatrix} = 0$$ Eventually, the two consecutive kernels in a sequence will be the same – $$\ker(A-\lambda I)^k = \ker(A-\lambda I)^{k+1}$$
3. When we have found the first such kernel, we take a look at the one that it succeeded – $$\ker(A-\lambda I)^{k-1}$$. We must now choose such a vector $$x_0$$ that is in $$\ker(A-\lambda I)^k$$ but not in $$\ker(A-\lambda I)^{k-1}$$. For example if $$\ker(A-\lambda I)^{k-1} = \mathcal{L}\left( \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \right)$$ whereas $$\ker(A-\lambda I)=\mathbb{R}^3$$, we can choose $$x_0=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$
4. We now consider a sequence: $$\begin{gather} x_1 = (A-\lambda I) x_0 \\ x_2 = (A-\lambda I) x_1 \\ \vdots \end{gather}$$ Eventually one vector will be zero, $$x_n = 0$$
5. Concatenate the vectors up to $$x_{n-1}$$ together from right to left $$A_{\lambda} = [ x_{n-1}, x_{n-2}, \ldots, x_1, x_0 ]$$
6. Repeat the above steps for every eigenvalue of $$A$$. Then concatenate the matrices $$A_{\lambda}$$ corresponding to each eigenvalue together. $$P = [ A_{\lambda_1}, A_{\lambda_2}, \ldots, A_{\lambda_n} ]$$ This is the so-called Jordan basis, the change-of-basis matrix we need to calculate the Jordan form. The order of $$A_{\lambda}$$ matrices does not matter since Jordan normal form is only unique up to a permutation of Jordan blocks.
7. We need to calculate the inverse of $$P$$, usually by Gaussian ellimination.
8. We calculate the Jordan form by $$J = P^{-1} A P$$.