How to calculate Jordan’s normal form (the hard way)

  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 $$(A-\lambda I)^n
    x_1 \\
    x_2 \\
    x_3 \\
    \vdots \\
    \end{bmatrix} = 0$$ 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 $$\ker(A-\lambda I)^{k-1} = \mathcal{L}\left( \begin{bmatrix}
    1 \\
    1 \\
    1 \\
    0 \\
    \end{bmatrix} \right)$$ whereas \(\ker(A-\lambda I)=\mathbb{R}^3\), we can choose $$x_0=\begin{bmatrix}
    0 \\
    0 \\
  4. We now consider a sequence: $$\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 \(J = P^{-1} A P\).

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