Let $A=\begin{pmatrix}2 && 2 && 3 \\ 1 && 3 && 3\\ -1 && -2 && -2\end{pmatrix}$

Finding the eigenvalues produce that $\lambda_1=\lambda_2=\lambda_3=1$

The geometry of $A$ is $2$, therefore exists only two blocks, and therefore the only possibility is $J_1(1),J_2(1)$.

This produce the following Jordan matrix: $J=\begin{pmatrix}1 && 1 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{pmatrix}$

Now choosing one basis of kernel of $A-I$, Let $B_1=\{v_1=\begin{pmatrix}3 \\ 3 \\ -3\end{pmatrix},v_2=\begin{pmatrix}0\\1\\-2/3\end{pmatrix}\}$ be such base (you can check it's basis, it's linearly independence and applying both vectors to $A-I$ gives $0$.

We can find the Jordan form, since solutions to $(A-I)w=v_1$ exist

However choosing $B_2=\{u_1=\begin{pmatrix}2 \\ -1 \\ 0\end{pmatrix},u_2=\begin{pmatrix}3 \\0\\-1\end{pmatrix}\}$, solution to $(A-I)w=u_1$ or $(A-I)w=u_2$ don't exist.

How can I find the Jordan form in the 2nd base? how come choosing one basis allows Jordan form to exist (or at-least allows for chaining of generalized eigenvectors) while the other doesn't exist?


$\begin{pmatrix}1 && 2 && 3 \\ 1 && 2 && 3\\ -1 && -2 && -3\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}2 \\ -1 \\ 0\end{pmatrix}$ gives no solution since it requires $x+2y+3z$ to have two distinct solutions.

$\begin{pmatrix}1 && 2 && 3 \\ 1 && 2 && 3\\ -1 && -2 && -3\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}3 \\ 0 \\ -1\end{pmatrix}$ gives no solution since it requires $x+2y+3z$ to have two distinct solutions.

Therefore if I wanted to chain using $B_2$, no solutions exist, however it's possible to chain using $B_1$. The question is, why does this happen? how can I ensure that I find a generalized eigenvector where this does not happen.

  • $\begingroup$ @Moo Oh, sorry; the last row is $-1, -2, -2$ I'll fix it in the post, the matrix isn't diagonalizable. But it does have a Jordan form; but apparently it does have a Jordan form for one basis of ker $A-I$ but not the other. $\endgroup$ – Rab Jun 11 '17 at 20:23
  • $\begingroup$ @Moo That's what I did, and I did solve it but the problem is that chaining with a one set of generalized eigenvectors $(B_1)$ gives a (valid) solution , and choosing another set of generalized eigenvectors $(B_2)$ gives no solution. and I am wondering why is that. $\endgroup$ – Rab Jun 11 '17 at 20:37
  • $\begingroup$ As such, all of $u_1, u_2, v_1,v_2$ are generalized eigenvectors, but you can chain with $B_1$ but not $B_2$ $\endgroup$ – Rab Jun 11 '17 at 20:39

The procedure for finding the basis is to start from a generalized eigenvector, that is, a vector $v_2$ such that $(A-I)^2v_2=0$, but $(A-I)v_2\ne 0$. Since $(A-I)^2=0$, you can take any vector in $\Bbb{R}^3\setminus Ker(A-I)$. Then $v_1=(A-I)v_2$ will be an eigenvector of $A$.

Note that $Im(A-I)$ is one-dimensional generated by $\begin{pmatrix} 1\\ 1\\ -1\end{pmatrix}$, so $v_1$ will always be a multiple of $\begin{pmatrix} 1\\ 1\\ -1\end{pmatrix}$.

Note also that $v_2$ cannot be in $Ker(A-I)$, since in that case $(A-I)v_2=0$.

In order to complete the basis, take another vector $v_3$ in $Ker(A-I)$ which is not in the subspace generated by $v_1$.

If you want to search directly for $v_1,v_2$ with $v_1=(A-I)v_2$, you have to note that $v_1$ is in the image of $A-I$, which is one dimensional, so it is a multiple of $\begin{pmatrix} 1\\ 1\\ -1\end{pmatrix}$.

So in your example you can take $v_2=\begin{pmatrix} 1\\ 0\\ 0\end{pmatrix}$, then $v_1=(A-I)v_2=\begin{pmatrix} 1\\ 1\\ -1\end{pmatrix}$, and take $v_3$ another vector of $Ker(A-I)$, for example $v_3=\begin{pmatrix} 2\\ -1\\ 0\end{pmatrix}$. Then the matrix of $A$ with respect to the basis $\{v_1,v_2,v_3\}$ will be $\begin{pmatrix} 1&1&0\\ 0&1&0\\0&0& 1\end{pmatrix}$, since $A(v_1)=v_1$, $A(v_2)=v_1+v_2$ and $Av_3=v_3$.


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