# Find Jordan normal form and basis

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.

• 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? – Greg Martin Mar 16 at 18:41
• @GregMartin I edited my post and I explained why I know that I have a mistake. – MP3129 Mar 16 at 19:15

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 $$-2$$).
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 $$\overset{A}{\left[\begin{array}{rrr} 0 & 1 & 1 \\ -4 & -4 & -2 \\ 0 & 0 & -2 \end{array}\right]} = \overset{P}{\left[\begin{array}{rrr} 2 & 1 & 0 \\ -4 & 0 & 1 \\ 0 & 0 & -1 \end{array}\right]} \overset{J}{\left[\begin{array}{rr|r} -2 & 1 & 0 \\ 0 & -2 & 0 \\ \hline 0 & 0 & -2 \end{array}\right]} \overset{P^{-1}}{\left[\begin{array}{rrr} 0 & -\frac{1}{4} & -\frac{1}{4} \\ 1 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & -1 \end{array}\right]}$$