find the Jordan basis of a matrix

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 $$i$$?

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

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: $$J=\begin{bmatrix} 5&0&0\\0&5&1\\0&0&5\end{bmatrix}.$$

• Why in the world did someone downvote this? – Moo Jan 1 at 1:54
• I understand why v3 must be in Ker(A-5I)^2, but why mustn't v3 be in E5? – Sam.S Jan 1 at 15:42
• 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. – Bernard Jan 1 at 16:08

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.

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

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