Jordan Normal Form using generalized eigenvector

Question

How do I find the Jordan Normal Form for this matrix $A=\begin{bmatrix} -2 & -3 & 6 \\ 1 & 2 & -2\\ -1 & -1 &3 \end{bmatrix}$ ?

This is my method but I am not able to get to the final answer:

There is one eigenvalue $\lambda=1$ with algebraic multiplicity 3.

$\ker(A-I_3)= \operatorname{span}\left\{\begin{pmatrix} -1\\1\\0\end{pmatrix},\begin{pmatrix} 2\\0\\1\end{pmatrix}\right\}$

To calculate the generalized eigenvector, we have $(A-I)^2v=0$, but since $(A-I)^2 = 0$ we pick an arbitrary $v_3$ that is independent to the other two vectors, such as $\begin{pmatrix} 1\\0\\0\end{pmatrix}$

So, I combined the 3 vectors and I now have the matrix $P=\begin{pmatrix} -1 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$

Then, using the formula $P^{-1} A P$, the result is $\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix} $, which is not in JNF.

Would anyone please guide me as to how to correctly find the JNF for this matrix?

Answer 1

You have indeed found generalized eigenvectors, but the your generalized eigenvector $v$ does not have the correct relationship with the eigenvectors that you selected.

The generalized eigenvectors used to produce the Jordan form of a matrix must consist of separate Jordan chains, with each chain corresponding to a single Jordan block. In this particular case, the Jordan form is given by $$ J = \pmatrix{1&1&0\\0&1&0\\0&0&1}. $$ The first Jordan block (of size $2$) corresponds to one Jordan chain, which is to say a pair of vectors $x,v$ such that $(A-I)x = 0$ and $(A-I)v = x$. The second Jordan block (of size $1$) corresponds to a single eigenvector that is linearly independent from the other two vectors.

As you have correctly noted, the generalized eigenvector $v$ must satisfy $(A - I)^2 v = 0$, but it also must satisfy $(A - I) v \neq 0$. In order to choose a suitable generalized $v$ and associated eigenvector $x$, it helps to look at $A - I$ (or more generally, the matrix $(A -\lambda I)^k$ where $(A - \lambda I)^k \neq 0$ but $(A - \lambda I)^{k+1} = 0$). The columns of $A - I$ have the following important properties:

• Each column $x$ satisfies $(A - I)x = 0$
• Each column $x$ is of the form $(A - I)v$ for some vector $v$
• A vector $v$ such that $(A - I)v = x$ satisfies $(A - I)^2v = (A - I)x = 0$

With that in mind, we see that $$ A - I = \pmatrix{ -3&-3&6\\ 1 & 1 & -2\\ -1& -1 & 2}. $$ The first column of $A$ is non-zero, so it is a suitable choice of $x$. Moreover, because $x = (-3,1,-1)$ is the first column of $A - I$, we can write $x$ in the form $x = A \mathbf e_1$, where $\mathbf e_1$ denotes the first column of the identity matrix, $\mathbf e_1 = (1,0,0)$. That is, $\mathbf e_1$ is a suitable choice for our generalized eigenvector $v$.

From there, we need to extend the set $\{x\}$ into a full basis for the eigenspace of $A$ associated with $1$. As you correctly computed, this eigenspace has the basis $\{(-1,1,0),(2,0,1)\}$. In this case, we can just use the first vector from the basis you computed: the set $\{(-3,1,-1),(-1,1,0)\}$ is a set of eigenvectors that is linearly independent and has two elements, so it is a basis of the eigenspace.

So, we can take $y = (-3,1,-1)$ and use $(x,v,y)$ as our Jordan basis.