# Determining the Jordan decomposition of a given matrix

$$\newcommand{\m}{\left( \begin{matrix} #1 \end{matrix} \right)} \newcommand{\l}{\lambda}$$ I have the matrix given as follows:

$$A := \m{ -2 & -1 & 1 & 2 \\ 1 & -4 & 1 & 2 \\ 0 & 0 & -5 & 4 \\ 0 & 0 & -1 & -1 }$$

I wish to determine the Jordan decomposition for $$A$$. So far I have determined the following:

• $$A$$ has characteristic polynomial $$p(\l) = -(\l+3)^4$$. (Wolfram agrees up to that leading coefficient which isn't relevant.) Thus, it has a single eigenvalue, $$\l = -3$$, with multiplicity $$4$$.

• The matrix $$A+3I$$ has nullity $$2$$. Thus the algebraic and geometric multiplicities of $$3$$ differ, making it not diagonalizable in the traditional sense. (Wolfram agrees.)

• I have found that there exist two ordinary eigenvectors for $$A$$, i.e. vectors $$\vec p$$ such that $$(A+3I)\vec p = \vec 0$$. Namely, the following: $$v = (-4,0,2,1) \qquad w = (1,1,0,0)$$ (Wolfram agrees with this being the basis of the null space.)

• I still need two generalized eigenvectors to fill my roster. So let's start with squaring $$A+3I$$. In such a case, there is a single vector, by my eyes, which fits. Using the alternate characterization of generalizing a specific eigenvector, e.g. finding $$v'$$ such that $$(A+3I)v' = v$$ (or $$w$$), we can see that $$v$$ won't work in this way, but if we do it for $$w$$ we can obtain $$w' = (1,0,0,0)$$ Wolfram does list this in the basis of the null space of $$(A+3I)^2$$. There isn't an issue picking $$w'$$ this way so far as far as I know: Wolfram agrees that the three are linearly independent. Thus, this is the only additional eigenvector we get from $$(A+3I)^2$$.

• To find a fourth (generalized) eigenvector, $$w''$$, we will note that $$(A+3I)^3$$ is the zero matrix and thus has null space of $$\Bbb R^4$$. (Again, Wolfram agrees.) Thus, we can pick any vector we please that is linearly independent from the rest. As it happens, $$w'' = (0,0,1,1)$$ is such a vector. (Again, Wolfram agrees.)

• From the structure of this problem, we see two Jordan chains: $$v \qquad w \to w' \to w''$$ Thus, the Jordan form $$J$$ of $$A$$ should have one $$1 \times 1$$ block and one $$3 \times 3$$ block, each associated with $$\l = -3$$. Thus, we will have $$J = \m{ -3 & 0 & 0 & 0\\ 0 & -3 & 1 & 0 \\ 0 & 0 & -3 & 1 \\ 0 & 0 & 0 & -3 }$$

• Thus, we have our four eigenvectors to form the matrix $$P$$ such that $$A = PJP^{-1}$$. $$P := \big[ \; v \; \big| \; w \; \big| \; w' \; \big| \; w'' \; \big] = \m{ -4 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 2 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1}$$

This is where I get confused. Performing the calculation of $$PJP^{-1}$$ does not give me quite $$A$$ back. Per Wolfram,

$$PJP^{-1} = \m{ -2 & -1 & 3 & -2 \\ 1 & -4 & 4 & -4 \\ 0 & 0 & -3 & 0 \\ 0 & 0 & 0 & -3}$$

yet I'm not entirely sure where I've gone wrong. I've been trying to resolve this discrepancy for days with no luck, no matter what I look up. In fact, Wolfram gives a very different decomposition: namely, they claim

$$P = \m{1/4 & 8 & 2 & 0\\ 5/4 & 8 & 2 & 0\\ 1/2 & 0 & 4 & 0\\ 1/4 & 0 & 2 & 1}$$

but at least we agree on $$J$$...

Can anyone enlighten me as to what I might have done wrong?

I like going from the ground up; the fractions wait until $$P^{-1},$$ where $$J = P^{-1} A P.$$ Here $$(A + 3 I)^3 = 0$$ but $$(A+3I)^2 \neq 0.$$ We need three column vectors $$p,q,r$$ such that $$(A+3I)^3 r = 0$$ but $$(A+3I)^2 r \neq 0.$$ I took $$r = (0,0,0,1)^T .$$ Then $$q = (A+3I) r = (2,2,4,2)^T$$ and $$p = (A+3I) q = (8,8,0,0)^T .$$ Thus $$p$$ is a genuine eigenvector
The first column of $$P$$ is another genuine eigenvector