given a Matrix or function, we been told to find jordan form of it. once we have found char poly, e-vals. We need to find e-vectors. Suppose we found them by rref or how gilbert strang shows us. my question is will we there be always be a solution that is $(A-I\lambda)^p x=0$ that is $(A-I\lambda)^{p-1} x $ is a vanilla e-vector. for example will there always be a solution by doing the inverse of each recursively like if p=3 the dot diagram would be like $$\begin{aligned} *(A-I\lambda)^2 v_1 \\ *(A-I\lambda)(v_1) \\ *(v_1) \end{aligned}$$ will this method always yeild a $v_1$
I recall one time I could not find the $v_1$ by recursively finding the inverses there was an instance where there was no solution. I had to change the eigenvectors to some wierd combination that was not the original how strang finds the e-vectors but I cannot recreate it.
My linear algebra teacher says that there always be a $v_1$ doing it the way of finding the inverse which I do not think so. I think there is times where there is no solution. And says that there it is always jordan form ( do think so.)
here is the sample I was talking about.
$$ A= \begin{pmatrix} 2& -1& 0& 1 \\ 0 & 3& -1& 0 \\ 0& 1& 1& 0 \\ 0& -1& 0 &3 \end{pmatrix}$$
this is ex 2 on 7.2 of linear algebra by Friedberg, Insel, Spence but I am finding e-vectors like I learned in linear one. the char poly is $(t-2)^3 (t-3)$. so
$\lambda_1 =2$ with a.m 3
$\lambda_2 =3 $ with mult 1
We will find eigenvector of $ \lambda_1$
$$ \begin{aligned} A-\lambda_1 I &= \begin{pmatrix} 2-2& -1& 0& 1 \\ 0 & 3-2& -1& 0 \\ 0& 1& 1-2& 0 \\ 0& -1& 0 &3-2 \end{pmatrix} \\ &= \begin{pmatrix} 0& -1& 0& 1 \\ 0 & 1 & -1& 0 \\ 0& 1& -1& 0 \\ 0& -1& 0 &1 \end{pmatrix} \end{aligned}$$ get that $$rref(A-\lambda_1 I)= \begin{pmatrix} 0& 1& 0& -1 \\ 0 & 0 & 1& -1 \\ 0& 0& 0& 0 \\ 0& 0& 0 &0 \end{pmatrix} $$ $x_1 =1,x_4=0 \Rightarrow (1,0,0,0)$ and $x_4=1,x_1=0 \Rightarrow (0,1,1,1)$. Those are our 2 e-vectors
Doing the dot diagram we get that
$$\begin{matrix} * (T-I\lambda_1)v_1 & *v_2
\\ *v_1 & \end{matrix} $$
now assume $(T-\lambda I)v_1=(1,0,0,0)$ and $v_2 = (0,1,1,1)$
we need to find $v_1$ that is $$\begin{pmatrix} 0 & -1& 0& 1 \\ 0 & 1 &-1& 0 \\ 0& 1 &-1 &0 \\ 0&-1&0&1 \end{pmatrix} v_1 = \begin{pmatrix}1 \\0 \\0\\0\\0 \end{pmatrix}$$
so $$ rref \begin{pmatrix} 0 &-1& 0& 1& |&1 \\ 0 &1& -1& 0& |& 0 \\ 0 &1& -1& 0& |& 0 \\0 &-1& 0& 1& |& 0\end{pmatrix} =\begin{pmatrix} 0 &1& 0& -1& & | &-1 \\ 0 &0& 1& -1& & |&-1 \\ 0& 0& 0& 0& & |&0 \\0& 0& 0& 0& & | &-1 \end{pmatrix}$$ No solution. maybe $v_1 = (0,1,1,1)$ so,
$$rref \begin{pmatrix} 0 &-1& 0& 1& |& 0 \\ 0 &1& -1& 0& |& 1 \\ 0 &1& -1& 0& |& 1 \\0 &-1& 0& 1& |& 1 \end{pmatrix} = \begin{pmatrix} 0 &1& 0& -1& |& 0 \\ 0 &0& 1& -1& |& 0 \\ 0 &0& 0& 0& |& 1 \\0 &0& 0& 0& |& 0 \end{pmatrix} $$ also no solution. So this method does not work all the time to find basis jordan form . I am being told that this method will always work.
Did I do something wrong with counter example?? I do want to email him this. I am not saying there is no jordan form just that this way does not work all the time just sometimes as long as started with e-vectors that work
AP = PJ
$(A - 2I)^2v = 0$
$\begin{pmatrix}0&-2&1&1\\0&0&0&0\\0&0&0&0\\0&-2&1&1\end{pmatrix}\begin{pmatrix}0\\1\\0\\2\end{pmatrix}$
Av = $\begin{pmatrix}3&2&2&1\\ 0&3&-1&0\\ 0&1&1&0\\ 0&-1&0&3\end{pmatrix}\begin{pmatrix}0\\1\\0\\2\end{pmatrix}=\begin{pmatrix}1\\1\\1\\1\end{pmatrix}$
$AP = PJ$
$\begin{pmatrix}3&2&2&1\\ 0&3&-1&0\\ 0&1&1&0\\ 0&-1&0&3\end{pmatrix}\begin{pmatrix}1&1&1&0\\ 0&0&1&1\\ 0&0&1&0\\ 1&0&1&2\end{pmatrix}=\begin{pmatrix}1&1&1&0\\ 0&0&1&1\\ 0&0&1&0\\ 1&0&1&2\end{pmatrix}\begin{pmatrix}3&0&0&0\\0&2&0&0\\0&0&2&1\\0&0&0&2\end{pmatrix}$