Finding Jordan basis of a matrix $(4\times 4)$

Question

I'm facing a problem finding a Jordan basis for this ($4 \times 4$) matrix: $$\left(\begin{matrix}3&-1&1&7\\9&-3&-7&-1\\0&0&4&-8\\0&0&2&-4\end{matrix}\right)$$

I know that the characteristic polynomial is $\lambda^4=0$ and it's minimal polynomial is: $x^2$. This means that $(A-0I)^2=0$. Also the Jordan form for this matrix is $$\left(\begin{matrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{matrix}\right)$$ But I'm facing problems in the general procedure to compute the basis for such a matrix?

Answer 1

Thank you everyone for the tips, i manage to solve it. The basis is

$$\left(\begin{matrix}-7 & 0 & 3 & 1\\-1 & 0 & 9 & 0\\-8 & 0 & 0 & 0\\-4 & 1 & 0 & 0 \end{matrix}\right)$$ And $P^{-1}AP=J$

Answer 2

Since:

$J$ is obtained by $A$ from a similarity transfomation $$P^{-1}AP=J$$

$P$ can be found solving the system $$AP=PJ$$

Answer 3

First, the minimal polynomial is not $\lambda^2$, and it seems to me that Ben's row reduction is wrong.

We have that the characteristic polynomial is $p(\lambda)=\lambda^4$, so, we already know that the jordan matrix will only have the value $0$ on the main diagonal.

Now we want to find the kernel of $(A-0I)=A$, so, reducing matrix $A$, we get

$$\pmatrix{ 3&-1&1&7\\ 9&-3&-7&-1\\ 0&0&4&-8\\ 0&0&2&-4} \leadsto \pmatrix{ 3&-1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0} $$

Solving the system is not difficult, $$ \pmatrix{ 3&-1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0} \cdot \pmatrix{ x_1\\ x_2\\ x_3\\ x_4} = \pmatrix{ 0\\ 0\\ 0\\ 0} $$ we get, $x_3=x_4=0$ and $3x_1=x_2$, if $x_1=1$, then $x_2=3$, then the base of the kernel is given by the vector, $(1,3,0,0)$. From this, we conclude that the form of jordan has only one block, so it is of the form,

$$\left(\begin{matrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{matrix}\right)$$

as the exponent of the minimal polynomial can be given by the size of the largest block, we have that the minimal polynomial is $p_m(\lambda)=\lambda^4$

You can go on and find the Jordan base, as follows:

You want to find a vector $v$, so that, $A^3v,A^2v,Av,v$, they are all non-null. Calculating the exponents, we obtain,

$$A^2=\pmatrix{ 0&0&28&-14\\ 0&0&0&126\\ 0&0&0&0\\ 0&0&0&0} \qquad A^3=\pmatrix{ 0&0&84&-168\\ 0&0&252&-504\\ 0&0&0&0\\ 0&0&0&0} $$

note that $e_3$ and $e_4$ are suitable vectors for what we want (because the third and fourth columns do not cancel and $A^4$ is the null matrix), let's choose $e_4$. So,

$$A^3e_4=(-168,-504,0,0)\qquad A^2e_4=(-14,126,0,0)\qquad Ae_4=(7,-1,-8,-4)$$

this way a Jordan base is

$$\{(0,0,0,1),(7,-1,-8,-4),(-14,126,0,0),(-168,-504,0,0)\}$$

it is not a pretty base, but this is a way to resolve the exercise quickly. We have so,

$$P= \pmatrix{ 0&7&-14&-168\\ 0&-1&126&-504\\ 0&-8&0&0\\ 1&-4&0&0} $$

And you can see that this solves the problem, $J=P^{-1}AP$.