4
$\begingroup$

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 :)

$\endgroup$

1 Answer 1

1
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ The (3,3)-entry of the Jordan matrix is -1, not 1. $\endgroup$
    – kodkod
    Sep 22, 2018 at 13:05
  • $\begingroup$ @kodkod thanks! $\endgroup$
    – Kenny Wong
    Sep 24, 2018 at 8:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .