I'm trying to follow an algorithm for finding the Jordan form for the matrix: $$ \begin{pmatrix} 0 & -1 \\ 4 & 4 \\ \end{pmatrix} $$ Its eigenvalue is 2 with algebraic multyplicity of 2. $$ A-2I = \begin{pmatrix} -2 & -1 \\ 4 & 2 \\ \end{pmatrix} $$ therefore $ker(A-2I)=sp\begin{pmatrix} 1\\-2 \end{pmatrix}$, and $dim(ker(A-2I))=1$. Next: $$ (A-2I)^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix} $$ so $dim(ker(A-2I)^n)=2$ for $n\geq2$. The next step is to find a vector $u_1$ in $ker(A-2I)^2 \setminus ker(A-2I)$. Problem is that for every $u_1$ (unless I made some dumb arithmetic error along the way, which I believe I haven't) I get that $(A-2I)u_1 = \alpha\begin{pmatrix} 1\\-2 \end{pmatrix}$ for some $\alpha\in\mathbb R$. e.g. $$(A-2I)e_1 = \begin{pmatrix} -2\\4 \end{pmatrix}$$

Okay, so I got a 1x1 Jordan block, and I'm left to find another. That means I ought to find some $m_2$ for which $$ m_2 = max \left\{ m : ker(A-2I)^m \setminus ker(A-2I)^{m-1} \cup sp\begin{pmatrix} 1\\-2 \end{pmatrix}\right\}$$ but since the spaces spanned by my eigenvector and $(A-2I)u_1$ are the same, the only m that can work is 0, that is if I interpret $ker(A-2I)^-1$ as an empty set or something.

How should I interpret it? And is that the correct choice for m?

  • $\begingroup$ If $\text{dim}(\operatorname{ker}(A - 2I)) = 1$ and $A$ is $2\times 2$ matrix, you cannot have a $1\times 1$ Jordan block. Also, $(A - 2I)e_1 = \alpha [1, -2]^T$ is expected. Your block is $2\times 2$ and your "Jordan basis" can be, for example $\{e_1, (A-2I)e_1\}$. $\endgroup$ – Antoine May 11 '16 at 8:45
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    $\begingroup$ Since $A$ has an eigenvalue with an algebraic multiplicity of $2$ but geometric multiplicity of $1$, the only possible Jordan form is $\pmatrix{2 & 1\\0&2}$. $\endgroup$ – Travis Willse May 11 '16 at 18:08

There is a theorem in Friedberg's linear algebra book that makes computing the Jordan form easy and algorithmic. We will construct a "dot diagram" for each eigenvalue that will tell us the size of the jordan blocks corresponding to that eigenvalue.

Let $\lambda$ be an eigenvalue of a linear transformation $T$. Compute the numbers $r_i$ defined as follows:

\begin{align} r_1 &= dim(V) - rank(T-\lambda I)\\ r_i &= rank((T-\lambda I)^{i-1}) - rank((T-\lambda I)^{i}) \end{align}

Now create a left-justified figure with rows and columns of dots. In the $i^{th}$ row, put $r_i$ dots. Once done, each column corresponds to a Jordan block! The number of dots in that column tells you the size of that block.

In this problem the only eigenvalue is 2. So we create the dot diagram corresponding to 2: $r_1 = 1$ and $r_2 = 1$. ($r_3 = 0$ and you can stop once you get zero). So we create a dot diagram with 1 dot in the first row and 1 dot in the second row.

\begin{matrix} \cdot \\ \cdot \end{matrix}

This has one column with 2 dots, so there is a Jordan block of size 2.

  • $\begingroup$ We were also explained the quicker way to get it done, but in this question we were asked to explicitly find the Jordan basis. Thanks! $\endgroup$ – Yoni May 13 '16 at 20:45

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