Finding a Jordan normal form and basis for matrix

I want to find a Jordan normal form and bases for matrix:

$A = \begin{pmatrix} 1&1&-1\\-3&-3&3\\-2&-2&2 \end{pmatrix}$

But it's characteristic polynomial is -$x^3$ which root is equal to 0.

Along with this result I see that this matrix do have a Jordan form: https://www.wolframalpha.com/input/?i=jordan+normal+form+calculator&rawformassumption=%7B%22F%22,+%22JordanDecompositionCalculator%22,+%22theMatrix%22%7D+-%3E%22%7B%7B1,+1,+-1%7D,+%7B-3,+-3,+3%7D,+%7B-2,+-2,+2%7D%7D%22

What should I do to find J matrix?

Thanks!

The standard method consists, for each eigenvalue $\lambda$, in considering the sequence $$\ker(A-\lambda I)\subset \ker(A-\lambda I)^2\subset \ker(A-\lambda I)^3\subset \dotsm$$ This sequence is first increasing, then stabilises. The stabilised subspace is the generalised eigenspace. If we denote $d_k=\dim \ker(A-\lambda I)^k$ and set $d_0=0$, the sequence $(d_k-d_{k-1})$ decreases to $0$, and the value of $d_k-d_{k-1}$ is equal to the number of Jordan blocks of size $\ge k$.

Here we have $d_1=2$ since $A$ has rank $1$. Hence there will be $2$ Jordan blocks, $d_2=3$ since $A^2=0$, hence there will be $1$ Jordan block of size $\ge 2$. Of course, we could find this result without this calculation, but I mention it for illustrative purposes.

Thus we arrive at $1$ Jordan block of size $1$ and $1$ Jordan block of size $2$. To find a Jordan basis, we take any vector in $\ker A^2\smallsetminus\ker A=\mathbf R^3\smallsetminus\ker A$, viz. $u_3=(0,0,1)$, then set $$u_2=Au_3=(-1,3,2)$$ Note $u_2$ is an eigenvector. Complete into a basis of $\ker A$ (the eigenspace) with, say $u_1=(1,0,1)$. The change of basis matrix will be $$P=\begin{bmatrix}1&-1&0\\0&3&0\\1&2&1\end{bmatrix},$$ its inverse is $$P^{-1}=\begin{bmatrix}1&\frac13&0\\0&\frac13&0\\-1&-1&1\end{bmatrix},$$ and the Jordan normal form of $A$ in this basis is $$\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}$$

Every Matrix has a Jordan Form! It might be a diagonal, but even that is a Jordan Normal Form!

To get a Jordan Normal Form, you need the eigenvalues $\lambda_i$, eigenvektors $v_i$ and it's generalized eigenvectors $u_i$.

You know, how to get the eigenvalues by solving $(A-\lambda_i I)v_i=0$ for $v_i$ but this will not give you a complete basis, only a subset. There are still eigenvalues with a multiplicity, that does not correspond to the eigenspace-dimension. Let this eigenvalue be $\lambda_k$ and it's eigenvektor $v_k$. You must now solve $(A-\lambda_k I)u_k = v_k$ in order to get the generalized eigenvektor. You repeat this (alsways having the newest generalized eigenvektor on the RHS) until you are done. This vektor $u_k$ satisfies $Au_k = \lambda_k u_k + v_k$. Therefor you will get $$A \begin{bmatrix}v_k &u_k\end{bmatrix} = \begin{bmatrix}v_k &u_k\end{bmatrix}\begin{bmatrix}\lambda_k&1 \\ 0 &\lambda_k\end{bmatrix}$$ From there you can see, how the Jordan normal Form is build.

• Every matrix with coefficients in an algebraically closed field. – Bernard Jan 17 '17 at 10:54
• Yes, you are right. I always forget about that since I just work on $\mathbb{C}$. – Laray Jan 17 '17 at 10:55

The characteristic polynomial of $A$ is $\chi(\lambda)=\lambda^3$. We have $A\ne 0$ and $A^2=0$ then, the minimum polynomial of $A$ is $\mu(\lambda)=\lambda^2$. According to a well known theorem, the Jordan normal form of $A$ is $$J=\begin{bmatrix}{0}&{1}&{}\\{0}&{0}&{}\\{}&{}&{0}\end{bmatrix}.$$