2
$\begingroup$

I'm trying to find the Jordan basis of the matrix $$A =\begin{bmatrix} 8 & 1 & 2 \\ -3 & 4 & -2\\ -3 & -1 & 3\end{bmatrix}$$ I've got the characteristic equation to be $CA(x) = (5-x)^3$ and hence the eigenvalue to be $5$. I started by finding $v_1$ such that $(A-5I)v_1=0$ and chose $v_1 = \begin{bmatrix} 1 \\ -3 \\ 0\end{bmatrix}$. I then tried to find a $v_2$ such that $(A-5I)^2v_2 = 0$ but $(A-5I)^2$ is the zero matrix so can $v_2$ be any vector in $\mathbb{R}^3$?

More generally, when trying to find a Jordan basis for a matrix $A$, what do you do when $(A-\lambda I)^i=0$ for some $i$? Also what do you do when there is no solution to $(A-\lambda I)^iv_i = 0$ for a particular $i$?

$\endgroup$
  • $\begingroup$ Please use MathJax to format your posts. $\endgroup$ – saulspatz Dec 31 '18 at 23:27
  • $\begingroup$ Ok, what is this? but for now, any ideas/help? $\endgroup$ – Sam.S Dec 31 '18 at 23:29
0
$\begingroup$

The eigenspace $E_5$ has dimension $2$ (so the Jordan form of the matrix will have $2$ Jordan blocks) since $$A-5I=\begin{bmatrix} 3&1&2\\-3&-1&-2\\-3&-1&-2\end{bmatrix} $$ and it is defined by the single equation $\;3x+y+2z=0$.

You should attack the problem backwards: begin with choosing a vector $v_3=(x,y,z)$, in $\ker(A-5I)^2\smallsetminus E_5=\mathbf R^3\smallsetminus E_5$, i.e. such that $$3x+y+2z\ne 0,\enspace \text{say }\enspace v_3=(1,0,-1),$$

and set $\;(A-5I)v_3=v_2=(1,-1,-1)$ ; $v_2$ belongs to the eigenspace $E_5$.

Last you have to complete $v_2$ with another vector $v_1$ in the eigenspace which is linearly independent from $v_2$. The vector you've found – $v_1=(1,-3,0)$ is fine. In the basis $\mathcal B=(v_1,v_2,v_3)$ the matrix of $A$ becomes by construction: $$J=\begin{bmatrix} 5&0&0\\0&5&1\\0&0&5\end{bmatrix}. $$

$\endgroup$
  • 1
    $\begingroup$ Why in the world did someone downvote this? $\endgroup$ – Moo Jan 1 at 1:54
  • $\begingroup$ I understand why v3 must be in Ker(A-5I)^2, but why mustn't v3 be in E5? $\endgroup$ – Sam.S Jan 1 at 15:42
  • 1
    $\begingroup$ This is because the algorithm starts with the generalised eigenvectors and ends with the real eigenvectorrs, obtained as images of the above level generalised eigenvectors – unless the matrix is diagonalisable. $\endgroup$ – Bernard Jan 1 at 16:08
2
$\begingroup$

If the minimal polynomial is $(\lambda -5)^2,$ you may, indeed, pick any column vector $w$ you like that is not already an eigenvector, make it the right hand column of $P. \; $ The rule is that the middle column must be $v = (A-5I) w.$ The left column of $P$ is then $u,$ a different eigenvector from $v.$ Sometimes it takes a little ingenuity to take the pair of eigenvectors you first calculated and revise to get $u.$

Then, with matrix $P,$ you get $J = P^{-1}AP.$

Try it.

$\endgroup$
  • $\begingroup$ Much appreciated. apologies for the lack of Mathjax. I will give this a go. $\endgroup$ – Sam.S Dec 31 '18 at 23:47
2
$\begingroup$

First take some $v \in \mathbb{C}^3$ such that $(A-5I)v \ne 0$, for example take $v = e_2$. Then calculate $$(A-5I)v = \begin{bmatrix} 1 \\ -1 \\ -1\end{bmatrix}$$ and find some $w \in \ker (A-5I)$ which is linearly independent with $(A-5I)v$. For example we can take the vector you already found: $$w = \begin{bmatrix} 1 \\ -3 \\ 0\end{bmatrix}$$

Then the Jordan basis is $\{(A-5I)v, v, w\} = \left\{\begin{bmatrix} 1 \\ -1 \\ -1\end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix} 1 \\ -3 \\ 0\end{bmatrix}\right\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.