# Archive for June, 2013

- 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 \(\lambda\)
- 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 – \(\ker(A-\lambda I)^k = \ker(A-\lambda I)^{k+1}\)
- 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}$$
- We now consider a sequence: $$\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$$
- Concatenate the vectors up to \(x_{n-1}\) together
**from right to left** $$A_{\lambda} = [ x_{n-1}, x_{n-2}, \ldots, x_1, x_0 ] $$
- 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.
- We need to calculate the inverse of \(P\), usually by Gaussian ellimination.
- We calculate the Jordan form by \(J = P^{-1} A P\).