Let $A\in \operatorname{End}(V)$ be a linear transformation of $V=\mathbf{C}^3$ which is represented by $$[A]=\begin{pmatrix} 1 &0 &0 \\ x_1 &2 &0 \\ x_2 &x_3 &1 \end{pmatrix}$$ w.r.t the standard basis, where $x_1,x_2,x_3\in \mathbf{C}$. The problem I am struggling with is, to determine the generalized eigenspaces and a Jordan normal form.

I found out that the eigenvectors of $[A]$ are $(0,1/x_3,1)$ and $(0,0,1)$ for eigenvalues $2$ and $1$, respectively. But I do not know how to determine the generalized eigenspaces. I checked it in Wikipedia, and it kind of confuses me as I am beginner to it.

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    $\begingroup$ It might be better to list the eigenvector as $(0,1,x_3)$ since that will work even in the case that $x_3=0.$ $\endgroup$ – ziggurism Oct 19 '16 at 19:22

For the eigenvalue $2$, since its algebraic multiplicity is $1$, so will be its geometric multiplicity. We have $$ \begin{bmatrix}2a\\ 2b\\ 2c\end{bmatrix} =\begin{bmatrix} 1 &0 &0 \\ x_1 &2 &0 \\ x_2 &x_3 &1 \end{bmatrix} \begin{bmatrix}a\\ b\\ c\end{bmatrix} =\begin{bmatrix} a\\ ax_1+2b\\ ax_2+bx_3+c \end{bmatrix} $$ From $2a=a$ we get $a=0$. Then the third equation becomes $bx_3+c=2c$, or $c=bx_3$. Thus the eigenvectors are of the form $$ \begin{bmatrix} 0\\ b\\ bx_3 \end{bmatrix} =b\,\begin{bmatrix} 0\\ 1\\ x_3 \end{bmatrix} $$ (note that I am not assuming that $x_3\ne0$, as you did).

For the eigenvalue $1$, the equations become $$ a=a, \ \ \ ax_1+2b=b, \ \ ax_2+bx_3+c=c. $$ It follows that $b=-ax_1$ and that $ax_2-ax_1x_3=0$. This last equation can be rewritten as $$\tag{*}a(x_2-x_1x_3)=0.$$

So we have two cases:

  • If $x_2=x_1x_3$, we get independent eigenvectors for the cases $a=0$ and $a\ne0$: $$ \begin{bmatrix} 0\\ 0\\1 \end{bmatrix},\ \ \ \text{ and }\begin{bmatrix}1\\-x_1\\ 0\end{bmatrix}. $$ In this case $A$ is diagonalizable and its Jordan form is $$\begin{bmatrix}1&0&0\\0&1&0\\ 0&0&2\end{bmatrix}.$$

  • When $x_2\ne x_1x_3$, the equation $(*)$ forces $a=0$, and we only get the eigenvectors $$ \begin{bmatrix} 0\\ 0\\c \end{bmatrix} =c\,\begin{bmatrix} 0\\ 0\\1 \end{bmatrix}. $$ In this case the Jordan form of $A$ will be $$\begin{bmatrix}1&1&0\\0&1&0\\ 0&0&2\end{bmatrix}. $$ For a generalized eigenvalue, we look at $(A-I)^2v=0$, i.e. $$ \begin{bmatrix} 0\\ 0\\0 \end{bmatrix}=\begin{bmatrix}0&0&0\\ x_1&1&0\\ x_1x_3&x_3&0\end{bmatrix}\begin{bmatrix} a\\ b\\c \end{bmatrix} =\begin{bmatrix} 0\\ ax_1+b\\ ax_1x_3+bx_3 \end{bmatrix} $$ As $ax_1x_3+bx_3=(ax_1+b)x_3=0$, the third entry gives us nothing. So $b=-ax_1$, and the generalized eigenvectors are of the form $$ \begin{bmatrix} a\\ -ax_1\\0 \end{bmatrix}. $$

  • $\begingroup$ The inequality in your second bullet should read $x_1x_3\neq x_2$? $\endgroup$ – ziggurism Oct 19 '16 at 19:54
  • $\begingroup$ Yes, thanks! $\ \ \ \ $ $\endgroup$ – Martin Argerami Oct 19 '16 at 19:58

The eigenspaces of $A$ are the kernels of $A-1$ and $A-2$. Since the multiplicity of eigenvalue 2 is one, it has no generalized eigenvalues.

For the eigenvectors with eigenvalue $1$, we find the kernel of $A-1$.

We want to solve

$$ \left(\begin{matrix} 0 & 0 & 0\\ x_1 & 2 & 0\\ x_2 & x_3 & 1 \end{matrix} \right) \left(\begin{matrix} v_1\\ v_2\\ v_3 \end{matrix} \right)=0 $$

So we get $x_1v_1+v_2=0$ and $x_2v_1+x_3v_2=0,$ and substitution gives $(x_2-x_1x_3)v_1=0.$

So I agree that $(0,0,1)$ is the only independent eigenvector in this kernel, but only if we assume $x_2-x_1x_3\neq 0$, so that we can solve the system with $v_1=v_2=0$.

On the other hand, if $x_2-x_1x_3=0$, then $(1,-x_1,0)$ and $(0,0,1)$ are both eigenvectors. In this case the matrix is diagonalizable, which also tells you its Jordan form.

So assuming that $x_2-x_1x_3\neq 0$, to find the generalized eigenvectors for eigenvalue 1, find the kernel of $(A-1)^2$. There is one independent vector there that is not also in the kernel of $A-1$ (so not an eigenvector), which is $(1, -x_1, 0)$.

In general to find the generalized eigenvectors for an eigenvalue $c$, you compute the kernel $(A-c)^k$, where $k$ is the power with which the $(x-c)$ factor occurs in the minimal polynomial.

Finally once you have the generalized eigenvector, you can write the Jordan canonical form by expressing the matrix in the ordered basis $(0,0,x_2-x_1x_3), (1,-x_1,0),(0,1,x_3)$. You get

$$ \left(\begin{matrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 2 \end{matrix} \right) $$


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