Jordan normal form of $\left(\begin{smallmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{smallmatrix}\right)$

Question

I want to find the Jordan normal form of $A=\begin{pmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{pmatrix}$, but somewhere I think that I make a mistake (I am quite new to the computation of the Jordan normal form).
I could find that $\lambda=3$ is $A$'s only eigenvalue and its algebraic multiplicity is $3$.
Afterwards, I solved the system $(A-3I_3)X=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ and I got that $x=\begin{pmatrix} -a-b \\ a \\ b \end{pmatrix}$ for some parametres $a, b\in \mathbb{C}$.
I picked the eigenvectors $X_1=\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ and $X_2=\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$ and then I got stuck. I tried solving both the equations $(A-3I_3)X'=X_1$ and $(A-3I_3)X''=X_2$, but these have no solutions, so I can't find any generalised eigenvectors. What should I do?

Answer 1

After you get the Jordan form (as psidaga's answer did) \begin{align*} J = \begin{pmatrix} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}, \end{align*} in order to determine the similarity matrix $P$ such that $P^{-1}AP = J$, it is is equivalent to solve the system \begin{align*} & A\alpha_1 = 3\alpha_1, \tag{1} \\ & A\alpha_2 = \alpha_1 + 3\alpha_2, \tag{2} \\ & A\alpha_3 = 3\alpha_3, \end{align*} where $\alpha_1, \alpha_2, \alpha_3$ are three columns of $P$. As you did, you can fix $\alpha_3$ to be $(-1, 0, 1)^T$. To determine $\alpha_1$ and $\alpha_2$, note that $A - 3I_{(3)}$ is nilpotent with nilpotent index $2$, i.e., $(A - 3I_{(3)})^2 = 0$, hence any non-zero vector $\alpha_2$ that is not already an eigenvector and $\alpha_1 := (A - 3I_{(3)})\alpha_2$ would satisfy equations $(1)$ and $(2)$, since $(A - 3I_{(3)})\alpha_1 = (A - 3I_{(3)})^2\alpha_2 = 0$. For example, let $\alpha_2 = (1, 0, 0)^T$, then $\alpha_1 = (1, -2, 1)^T$. It can be easily verified that \begin{align*} P = \begin{pmatrix} 1 & 1 & -1 \\ -2 & 0 & 0 \\ 1 & 0 & 1 \end{pmatrix} \end{align*} is invertible and satisfies $A = PJP^{-1}$.

Of course, $P$ is not unique, depending on what $\alpha_2$ you are choosing.

Answer 2

Since you have two linearly independent eigenvectors, you know that there are two blocks for $\lambda=3$. That is already enough to tell you what the Jordan form looks like.

So it is not really necessary to look for generalized eigenvectors, if you only want Jordan form. But still, it might be useful to try this - especially since you've mentioned problems with this part in your question.

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We want to have a solution for $(A-3I)X'=X$, where $X$ is an eigenvector. We do not know which eigenvector to chose for $X$, so let us try to do this with arbitrary eigenvector.

$\left(\begin{array}{ccc|c} 1 & 1 & 1 &-a-b \\ -2 &-2 &-2 & a \\ 1 & 1 & 1 & b \end{array}\right)\sim \left(\begin{array}{ccc|c} 0 & 0 & 0 &-a-2b \\ 0 & 0 & 0 & a+2b \\ 1 & 1 & 1 & b \end{array}\right)$

Already after these two steps we see, that this is only possible if $a+2b=0$. For example, if we choose $b=1$ and $a=-2$, then the above system is $$\left(\begin{array}{ccc|c} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{array}\right)$$ and one of the solutions is $(1,0,0)^T$. This is what we can choose for our generalized eigenvector.

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Another thing worth noticing is that if we want to have $(A-3I)X'=X$ then $X$ needs to be in the column space of the matrix $(A-3I)$. So we have $X$ in the span of $(1,-2,1)^T$. For example, we can choose $X'=\begin{pmatrix}1\\0\\0\end{pmatrix}$ and $X=\begin{pmatrix}1\\-2\\1\end{pmatrix}$.

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Once we found all eigenvectors and generalized eigenvectors, we can use them to get a matrix such that $A=PJP^{-1}$. So for $$P= \begin{pmatrix} 1 & 1 &-1 \\ -2 & 0 & 1 \\ 1 & 0 & 0 \\ \end{pmatrix} \qquad\text{and}\qquad J= \begin{pmatrix} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \\ \end{pmatrix} $$ we have $$A=PJP^{-1}.$$ One can check this by direct computation. Here is also a link to WolframAlpha.