# Jordan normal form (Basis)

Define $A = \begin{pmatrix} -7 & -32 & -32 & -35 \\ 1 & 5 & 4 & 4 \\ 1 & 4 & 5 & 5 \\ 0 & 0 & 0 & 1 \end{pmatrix} \in \mathbb{C^{4x4}}$

I computed the Jordan normal form of this matrix:

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

But how to find a Jordan basis for this matrix?

Since $ker(A-I)^3=0$ I can't choose a $v \in ker(A-I)^3$ \ $ker(A-I)^2$

Also, I tried to use $v \mapsto Av$

$Ab_1=b_1$,

$Ab_2=b_2$,

$Ab_3=b_2+b_3$,

$Ab_4=b_3+b_4$

But I don't know how to continue.

• i put an answer. Looking at what you wrote, it should be $(A-I)^3 = 0$ so $\ker (A-I)^3$ is all of $\mathbb C^4$ – Will Jagy May 18 '18 at 17:49

If you write down $(A-I)^2$ you see that any vector with the final entry nonzero is outside the kernel, so you can pick $$p_4 = \left( \begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right)$$ followed by $p_3 = (A-I) p_4 \;, \;$ $p_2 = (A-I) p_3 \;, \;$ which is automatically an eigenvector because $(A-I)p_2 = (A-I)^3 p_4 = 0.$ Then let the first column $p_1$ be a different eigenvector from $p_2.$ Notice that, with two eigenvectors available, I deliberately chose a linear combination so the entires of the first column would be small. I hoped it would reduce the determinant of $P,$ and this worked out nicely, the determinant came out to $2.$ You can see this is the factor of $\frac{1}{2}$ required to write all the matrices with integers.

$$\frac{1}{2} \left( \begin{array}{cccc} 1&5&3&0 \\ -9&-35&-35&0 \\ 2&8&8&0 \\ 0&0&0&2 \end{array} \right) \left( \begin{array}{cccc} 0&-8&-35&0 \\ 1&1&4&0 \\ -1&1&5&0 \\ 0&0&0&1 \end{array} \right) = \left( \begin{array}{cccc} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{array} \right)$$

$$\frac{1}{2} \left( \begin{array}{cccc} 1&5&3&0 \\ -9&-35&-35&0 \\ 2&8&8&0 \\ 0&0&0&2 \end{array} \right) \left( \begin{array}{cccc} -7&-32&-32&-35 \\ 1&5&4&4 \\ 1&4&5&5 \\ 0&0&0&1 \end{array} \right) \left( \begin{array}{cccc} 0&-8&-35&0 \\ 1&1&4&0 \\ -1&1&5&0 \\ 0&0&0&1 \end{array} \right) = \left( \begin{array}{cccc} 1&0&0&0 \\ 0&1&1&0 \\ 0&0&1&1 \\ 0&0&0&1 \end{array} \right)$$ and $$\frac{1}{2} \left( \begin{array}{cccc} 0&-8&-35&0 \\ 1&1&4&0 \\ -1&1&5&0 \\ 0&0&0&1 \end{array} \right) \left( \begin{array}{cccc} 1&0&0&0 \\ 0&1&1&0 \\ 0&0&1&1 \\ 0&0&0&1 \end{array} \right) \left( \begin{array}{cccc} 1&5&3&0 \\ -9&-35&-35&0 \\ 2&8&8&0 \\ 0&0&0&2 \end{array} \right) = \left( \begin{array}{cccc} -7&-32&-32&-35 \\ 1&5&4&4 \\ 1&4&5&5 \\ 0&0&0&1 \end{array} \right)$$

From here we can solve as follow

• $Ab_1=b_1\implies (A-I)b_1=0$ and find $b_1$

• $Ab_2=b_2\implies (A-I)b_2=0$ and find $b_2$

• $Ab_3=b_2+b_3 \implies (A-I)b_3=b_2$ and find $b_3$

• $Ab_4=b_3+b_4 \implies (A-I)b_4=b_3$ and find $b_4$