# Finding a Jordan basis of a $3\times 3$ matrix

Find a Jordan basis for the following matrix: $$A= \begin{pmatrix} 1 & -3 & 4 \\ 4 & -7 & 8 \\ 6 & -7 & 7 \\ \end{pmatrix}$$

Hey everyone. First I have found the characteristic polynomial which is $(x-3)(x+1)^2$.

Then i've found a basis for: $$\ker(3I-A)=\ker\begin{pmatrix}2 & 3 & -4 \\-4 & 10 & -8 \\-6 & 7 & -4 \\\end{pmatrix}$$ $B_1= \{(\frac{1}{2},1,1)^T\}$. Then, for the next Jordan block I've calculated $\ker(-I-A)=\ker \begin{pmatrix} -2 & 3 & -4 \\ -4 & 6 & -8 \\ -6 & 7 & -8 \\ \end{pmatrix}$ and found a basis for this subspace- $B_2= \{(1,2,1)^T\}. \dim(\ker(-I-A))=1\neq a_m(\lambda)=2$ so we find a basis for $\ker(-I-A)^2 \Rightarrow B_3=\{(1,1,0)^T,(0,1,1)^T\}$ and choose $e_1=(1,0,0)^T \ (e_1 \notin Sp(B_2))$ to complete $B_3$ to a basis of $\mathbb{R^3}$.

Hence, our first chain is the vector $v_1= \{(\frac{1}{2},1,1)^T\}$, and the second chain is $\{(-I-A)e_1, e_1\}=\{(-2,-4-6)^T, (1,0,0)^T\}$ therefor our basis is $B= \{(\frac{1}{2},1,1)^T, (-2,-4-6)^T, (1,0,0)^T \}$

But $P^{-1}AP$ where $P=\begin{pmatrix} \frac{1}{2} & -2 & 1 \\ 1 & -4 & 0 \\ 1 & -6 & 0 \\ \end{pmatrix}$ equals $\begin{pmatrix} 3 & -32 & 0 \\ 0 & -1 & -1 \\ 0 & 0 & -1 \\ \end{pmatrix} \neq \begin{pmatrix} 3 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \\ \end{pmatrix}$

I've done numerous tries with different vectors other than $e_1$ yet I did not achieve the matrix' Jordan form. I would be happy if you could help me find my mistakes. Thanks in advance :)

The problem is that $$e_1$$ must be chosen to be in $${\rm ker}(-I-A)^2$$. To be more specific, you must chose $$e_1$$ to be any element in $${\rm ker}(-I-A)^2$$ that is not in $${\rm ker}(-I-A)$$.

For example, you can chose $$e_1 = (1, 1, 0)^T.$$

Having chosen $$e_1$$, you are forced to take $$e_2 : = (A + I)e_1 = (-1, -2, -1)^T$$ as your generator of $${\rm Ker}(-I-A)$$.

Finally, you can take $$e_3 =(\tfrac 1 2, 1, 1)^T$$ as your generator of $${\rm Ker}(3I - A)$$.

So the vectors $$\{ e_1, e_2, e_3 \}$$ satisfy the relations

$$Ae_1 = -e_1 + e_2, \ \ \ A e_2 = -e_2, \ \ \ Ae_3 = 3e_3.$$

Thus $$\{ e_3, e_2, e_1 \}$$ is a Jordan basis, corresponding to the Jordan matrix

$$\begin{bmatrix} 3 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{bmatrix}$$

[I had to reorder the basis vectors to put the matrix in standard form - sorry about that!]

Finally, a couple of minor points:

• For the signs to work out, you need to take $$e_2 := (A + I)e_1$$ as your generator of $${\rm Ker}(-I-A)$$, not $$(-I - A)e_1$$.

• Verifying that $$e_2 = (A + I)e_1$$ really is in $${\rm Ker}(- I - A)$$ is a good sanity check!

• The (3,3)-entry of the Jordan matrix is -1, not 1. Sep 22, 2018 at 13:05
• @kodkod thanks! Sep 24, 2018 at 8:29