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Let $$A=M(\varphi)^{st}_{st}={\begin{bmatrix}0&1&1\\-4&-4&-2\\0&0&-2\end{bmatrix}}$$ and $ \varphi: \mathbb R^{3} \rightarrow \mathbb R^{3}$. Find the Jordan normal form $J_{A}$ for the matrix $A$ and a basis $X$ for the endomorphism $\varphi$ such that $J_{A}=M(\varphi)^{X}_{X}$.


I did this task but unfortunately my basis is not correct and I do not know where I'm making a mistake.

My try:

The appointment of $ J_ {A} $ is clear to me, so I will write only the result: $$J_{A}={\begin{bmatrix}-2&1&0\\0&-2&0\\0&0&-2\end{bmatrix}}$$Then I am trying to find basis:

$$(A+2I)^{2}={\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}}$$

$$\alpha_{2} \in \ker(\varphi+2id)^{2}- \ker(\varphi +2id)$$

$$\ker(\varphi+2id)^{2}=\mathbb R^{3}$$

$$\ker(\varphi+2id)=lin\left\{(1,-2,0),(0,-1,1)\right\}$$

From the above conclusions I think that I can take:

$$\alpha_{2}=(0,1,0)$$

$$\alpha_{1}=(\varphi+2id)(\alpha_{2})=(1,-2,0)$$

As $\alpha_{3}$ I choose a linearly independent vector and I have for example:

$$\alpha_{3}=(0,0,1)$$

So my basis is:

$$X=\left\{(1,-2,0),(0,1,0),(0,0,1)\right\}$$ I know that basis can be different. That's why I checked my answer:
$$J_{A}=M(\varphi)_{X}^{X}=M(id)^{X}_{st}M(\varphi)^{st}_{st}M(id)^{st}_{X}$$ From my basis I have: $$M(id)^{st}_{X}={\begin{bmatrix}1&0&0\\-2&0&1\\0&1&0\end{bmatrix}}$$Then I made a multiplication: $$M(id)^{X}_{st}M(\varphi)^{st}_{st}M(id)^{st}_{X}$$ and I get: $${\begin{bmatrix}-2&1&1\\0&-2&0\\0&0&-2\end{bmatrix}}$$Of course this is not equal $J_{A}$ so I know that I have a mistake.

I would like to add that it is very strange for me that when I was curiosity I changed the vector $\alpha_{1}$ and $\alpha_{2}$ then I get $J_{A}$. However by my calculation I can't do it.

Can you help me and say where I am doing something wrong? I suspect that I make some mistake in determining $ \alpha_{2} $ but I still do not realize what.

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  • $\begingroup$ Why do you believe that your answer is incorrect? In general, if someone tells you they've computed a basis for one of these normal form problems, do you know how to check whether they're answer is correct? Have you done this check on your own answer? $\endgroup$ – Greg Martin Mar 16 at 18:41
  • $\begingroup$ @GregMartin I edited my post and I explained why I know that I have a mistake. $\endgroup$ – MP3129 Mar 16 at 19:15
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It looks like you've successfully obtained the data of $A$ and its Jordan form $J$ as \begin{align*} A &= \left[\begin{array}{rrr} 0 & 1 & 1 \\ -4 & -4 & -2 \\ 0 & 0 & -2 \end{array}\right] & J &= \left[\begin{array}{rr|r} -2 & 1 & 0 \\ 0 & -2 & 0 \\ \hline 0 & 0 & -2 \end{array}\right] \end{align*} Note that our table of eigenvalues is $$ \begin{array}{c|c|c} \lambda & \operatorname{am}_A(\lambda) & \operatorname{gm}_A(\lambda) \\ \hline -2 & 3 & 2 \end{array} $$ Here, $\operatorname{am}_A(-2)$ is the algebraic multiplicity of $-2$ as an eigenvalue of $A$ (the number of times $-2$ appears on the diagonal of $J$) and $\operatorname{gm}_A(-2)$ is the geometric multiplicity (the number of Jordan blocks in $J$ corresponding to $-2$).

Let's find the change of basis matrix using this algorithm described by Stefan Friedl.

First, we compute the numbers $$ d_k = \operatorname{nullity}((A+2\cdot I)^k)-\operatorname{nullity}((A+2\cdot I)^{k-1}) $$ for $1\leq k\leq\operatorname{gm}_A(-2)=2$. Computing these numbers gives \begin{align*} d_1 &= 2 & d_2 &= 1 \end{align*} We use these numbers to construct a diagram $$ \begin{array}{cc} \Box & \Box \\ \Box \end{array} $$ According to our algorithm, we start at the bottom of this diagram and fill the boxes in row $k$ with linearly independent vectors that belong to $\operatorname{Null}((A+2\,I)^k)$ but not $\operatorname{Null}((A+2\,I)^{k-1})$. As soon as a box is filled with a vector $\vec{v}$, we fill the box above with $(A+2\,I)\vec{v}$.

By noting that \begin{align*} A+2\,I &= \left[\begin{array}{rrr} 2 & 1 & 1 \\ -4 & -2 & -2 \\ 0 & 0 & 0 \end{array}\right] & (A+2\,I)^2 &= \left[\begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{align*} we see that our diagram can take the form $$ \begin{array}{cc} \fbox{$\left\langle2,\,-4,\,0\right\rangle$} & \fbox{$\left\langle0,\,1,\,-1\right\rangle$}\\ \fbox{$\left\langle1,\,0,\,0\right\rangle$} \end{array} $$ This defines our change of basis matrix $P$ as $$ P=\left[\begin{array}{rrr} 2 & 1 & 0 \\ -4 & 0 & 1 \\ 0 & 0 & -1 \end{array}\right] $$ We can check our work by verifying that $A=PJP^{-1}$. Indeed, we have $$ \overset{A}{\left[\begin{array}{rrr} 0 & 1 & 1 \\ -4 & -4 & -2 \\ 0 & 0 & -2 \end{array}\right]} = \overset{P}{\left[\begin{array}{rrr} 2 & 1 & 0 \\ -4 & 0 & 1 \\ 0 & 0 & -1 \end{array}\right]} \overset{J}{\left[\begin{array}{rr|r} -2 & 1 & 0 \\ 0 & -2 & 0 \\ \hline 0 & 0 & -2 \end{array}\right]} \overset{P^{-1}}{\left[\begin{array}{rrr} 0 & -\frac{1}{4} & -\frac{1}{4} \\ 1 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & -1 \end{array}\right]} $$

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  • $\begingroup$ I really thank you for your work but I need also knowledge why my sollution has a mistake because I think it is good to know different methods but also what mistakes are made in your ideas. Do you have anything to say, what could have happened in my way? $\endgroup$ – MP3129 Mar 16 at 20:21
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    $\begingroup$ If you compare the two solutions, you should be able to see where our methods diverge. $\endgroup$ – Brian Fitzpatrick Mar 16 at 22:21

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