$A=\begin{bmatrix} -1 & 4 & -4 \\ 0 & 1 & 0 \\ 1 & -2 & 3 \end{bmatrix}$

The eigenvalues of the matrix are all $1$. The dimension of it's eigenspace is 2 so the Jordan normal form of the matrix is \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

this is all confirmed by WolframAlpha.

Now, an eigenvector for $1$ is $(0,1,1)$ but when I try t solve $AP=PJ$ where

$P=\begin{bmatrix} 0 & a & d \\ 1 & b & e \\ 1 & c & f \end{bmatrix}$

I get $1+b=b$ for the middle element. Why is $(0,1,1)$ a bad choice for the first column here, and how do I find the Jordan basis in this case?

  • $\begingroup$ @Amzoti Those eigenvectors then form a Jordan basis, correct? But if I already have one of them, then how can it be that my system that's supposed to find the other ones has no solutions? It would seem that it matters which column of the matrix $P$ I put the $(0,1,1)$ in, but how do I know the correct order? I guess what I'm asking is, if I have a Jordan basis $\{e_1, e_2, e_3\}$, how do I know in which order to put them to form the matrix $P$ that satisfies $AP=PJ$? $\endgroup$ Feb 5 '15 at 14:48

The characteristic polynomial is given by

$$\begin{align} p (\lambda) &= (-1-\lambda)(1 - \lambda)(3 - \lambda) - (-4)(1- \lambda)(1) \\&= (1 - \lambda)[-(-1-\lambda)(3-\lambda) - 4]\\&= -(\lambda - 1)^3\end{align}$$

Which gives you eingenvalue $\lambda = 1$ of multiplicity $3$. To find the eigenvectors

$$(A - \lambda I)\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix} = 0$$

then $u = \begin{pmatrix}2u_2 - 2u_3\\u_2\\u_3\end{pmatrix}$, then take $u = \begin{pmatrix}0\\1\\1\end{pmatrix}$ (your same idea) and $u' = \begin{pmatrix}2\\0\\-1\end{pmatrix} $ $L.I.$. Now to find a generalized vector we make

$$(A - I) v = u'$$

Then $v = \begin{pmatrix}-1 + 2v_2 - 2v_3\\v_2\\v_3\end{pmatrix}$, we may take $v = \begin{pmatrix}-1\\0\\0\end{pmatrix}$. A basis would then be


As we have two $L.I.$ eigenvectors there is going to be two Jordan blocks, a $1 \times 1$ block corresponding to $u$ and $2\times 2$ block corresponding to $u'$ and $v$. That gives us

$$J =\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}$$

  • $\begingroup$ This works well, I juse have one issue. The second eigenvector you picked is one of infinitely many vectors from the eigenspace that are linearly independent of $(0,1,1)$. Yet, it's the only one for which $(A-I)v=u$ has a solution. How do you, in general, know which one to pick? $\endgroup$ Feb 5 '15 at 17:37
  • $\begingroup$ Ok, not the only one. It's one of two because $(-2, 0, 1)$ would also work. $\endgroup$ Feb 5 '15 at 17:38
  • $\begingroup$ I was just typing this. You could also work multiples. $\endgroup$ Feb 5 '15 at 17:39
  • $\begingroup$ Still, you need a zero in the middle to be able to solve the system, yet this isn't a requirement for linear independence with the first vector. $\endgroup$ Feb 5 '15 at 17:40
  • $\begingroup$ If you think, you could've chosen a different $u$, there are infinitely many solutions, so just take one and try to work with it. The important thing is to analyze the rank and nullity of your matrix, in order to know how many jordan blocks you will have. $\endgroup$ Feb 5 '15 at 17:45

You have to find a basis of its eigenspace $\{e_1,e_2\}$ first, then a vector $e_3$ such that $(A-I)e_3=e_2$, which means $Ae_3=e_3+e_2$, hence wrt the basis $\{e_1,e_2,e_3\}$ the matrix has the form:$$\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}.$$

In practice: $(A-I)^2=0$, so $\,\operatorname{im}(A-I)\subset \ker (A-I)$. Take any vector not in the eigenspace as $e_3$, say $\,^{\mathrm t}(1,0,0)$, and $\,e_2=(A-I)e_3= {}^{\mathrm t}(-2,0,1)$; it is an eigenvector.

Complete with an eigenvector $e_1$ independent from $e_2$; since the equation of the eigenspace is $\,x=2(y-z)$, we can take $e_1={} ^{\mathrm t}(2,1,0)$.

In the basis $(e_1,e_2,e_3)$, we have $Ae_3=e_2+e_3$ (by construction), $Ae_2=e_2$, $Ae_1=e_1$, so the matrix of $A$ in that base has the aforementioned form.


  • $\begingroup$ So, are you saying that the mistake was my ordering of the Jordan blocks? I thought their order didn't matter. $\endgroup$ Feb 5 '15 at 13:15
  • $\begingroup$ Actually,I thought that $\{(0,1,1), (2,1,0)\}$ is a basis, but neither $(A-I)x=e_1$ nor $(A-I)x=e_2$ have a solution because $A-I$ has all zeroes in the middle row and neither of the basis vectors have a zero in the middle. What did I make a mistake? $\endgroup$ Feb 5 '15 at 13:24
  • $\begingroup$ @Luka Horvat: The order of the Jordan blocks has no importance (it depends on the numbering of the vectors in the basis), but the way I describe the construction of the basis in practice) ends up naturally in the form I describe. See my completed answer. You can't find a solution because the image of $A-I$ has dimension $1$, while the eigenspace has dimension $2$, so any eigenvector is not necessarily in this image. The solution consists in reversing the procedure: start with a vector not in the eigenspace, take its image, which is an eigenvector and complete the basis of the eigenspace. $\endgroup$
    – Bernard
    Feb 6 '15 at 1:23

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