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Let $ M= \begin{bmatrix} 0 & 0 & 0 \\ 1 & 1 &1 \\ -1 & -1 & -1\\ \end{bmatrix}$ $\in M_{3}(\mathbb R^{3})$

I found that $\chi_{M}=X^{3}$ and $\mu_{M}=X^{2}$, since $\dim_{\mathbb R}\operatorname{Im}(M-0\cdot\operatorname{Id})=1$

So $$\ker(M-0\cdot\operatorname{Id})= \ker\begin{bmatrix} 0 & 0 & 0 \\ 1 & 1 &1 \\ -1 & -1 & -1\\ \end{bmatrix}=\left\{\begin{bmatrix} -1\\ 1 \\ 0 \\ \end{bmatrix}, \begin{bmatrix} -1\\ 0 \\ 1 \\ \end{bmatrix} \right\}\quad(*)$$

Now define $(M-0\cdot\operatorname{Id})=N$.

It follows:

$0 \subset \ker N \subset \mathbb R^{3}$. We can find a basis of the complement to $(*)$ in $\mathbb R^{3}$, say $\begin{bmatrix} 1\\ 0 \\ 0 \\ \end{bmatrix}$. Now,

$N\begin{bmatrix} 1\\ 0 \\ 0 \\ \end{bmatrix}=\begin{bmatrix} 0\\ 1 \\ -1 \\ \end{bmatrix} \in \ker N$. Now we find an element in $\ker N$ such that $\{\begin{bmatrix} 0\\ 1 \\ -1 \\ \end{bmatrix}, v_{1}\}$ is a basis of $\ker N$. Let's choose $v_{1}=\begin{bmatrix} -2\\ 1 \\ 1 \\ \end{bmatrix}$. So our Jordan basis is $$\left\{\begin{bmatrix} 1\\ 0 \\ 0 \\ \end{bmatrix}, \begin{bmatrix} 0\\ 1 \\ -1 \\ \end{bmatrix}, \begin{bmatrix} -2\\ 1 \\ 1 \\ \end{bmatrix}\right\}$$

Problem: For $S=\begin{bmatrix} 1 & 0 & -2\\ 0 & 1 & 1\\ 0 & -1 & 1\\ \end{bmatrix}$ and $J=S^{-1}MS$ , I only get a Jordan Matrix $J$ with ones below the diagonal, I am aware that this is also the Jordan Form, nonetheless in our lectures we state that $J$ has ones above the diagonal. For peace of mind, I would like to know I am not getting ones above the diagonal but rather below the diagonal?

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  • $\begingroup$ So there is no sure way in which to ensure that a Jordan Basis has ones above the diagonal as opposed to below the diagonal? $\endgroup$ – SABOY Jun 21 '18 at 12:34
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You must construct the Jordan basis more carefully.

You already determined that the Jordan form consists of one $2 \times 2$ block and one $1 \times 1$ block, i.e. $\dim \ker M = 2$ and $\dim \ker M^2 = 3$.

Write a table like this $$ \begin{array}{ccc} \ker M^2 \,\dot-\,\ker M^1 & f_2\\ \ker M^1 \,\dot-\,\ker M^0 & f_1 & f_3\\ \end{array} $$

Pick a vector $f_2 \in \ker M^2 \setminus \ker M^1$, such as $f_2 = \pmatrix{1 \\ 0 \\ 0}$ and define $f_1 = Mf_2 = \pmatrix{0 \\ 1 \\ -1}$.

Pick a vector $f_3 \in \ker M$ linearly independent with $f_1$, such as $f_3 = \pmatrix{-1 \\ 1 \\ 0}$.

Now $$\{f_1, f_2, f_3\} = \left\{\pmatrix{0 \\ 1 \\ -1}, \pmatrix{1 \\ 0 \\ 0}, \pmatrix{-1 \\ 1 \\ 0}\right\}$$ is a Jordan basis for $M$, with the Jordan form above the diagonal. Indeed: $$Mf_1 = 0$$ $$Mf_2 = f_1$$ $$Mf_3 = 0$$

Put those vectors in a matrix $S$ and we have $$J = S^{-1}MS = \pmatrix{0 & 1 & -1 \\ 1 & 0 & 1 \\ -1 & 0 & 0}^{-1}\pmatrix{0 & 0 & 0 \\ 1 & 1 & 1 \\ -1 & -1 & -1}\pmatrix{0 & 1 & -1 \\ 1 & 0 & 1 \\ -1 & 0 & 0} = \pmatrix{0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0}$$

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  • $\begingroup$ So clearly the upper/lower diagonal of 1s is determined by my selection of $f_{3}$. How do I know which $f_{3}$ is the right one to select??? $\endgroup$ – SABOY Jun 21 '18 at 20:58
  • $\begingroup$ @SABOY You can pick any $f_3$ such that $\{f_1, f_3\}$ is a basis for $\ker M^1 \,\dot-\,\ker M^0 = \ker M$. $\endgroup$ – mechanodroid Jun 22 '18 at 9:16
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If your method has produced an $r\times r$ Jordan block $$\left[\matrix{\lambda&0&0&0&0\cr 1&\lambda&0&0&0\cr 0&1&\lambda&0&0\cr 0&0&1&\lambda&0\cr 0&0&0&1&\lambda\cr}\right]\ ,$$ valid for some basis vectors $e_1$, $\ldots$, $e_r$ (part of a larger basis), then replace the $e_i$ $(1\leq i\leq r)$ by $$f_i:=e_{r+1-i}\qquad(1\leq i\leq r)\ ,$$ and you obtain a Jordan block with the $1$s above the diagonal.

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