# Find Jordan canonical form and basis of a linear operator.

Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear operator such that: $T(x,y,z)=(-y-2z,x+3y+z,x+3z)$, I need to find a Jordan canonical form and a basis. This is what i did:

In the first place, I found the associated matrix to this linear operator in the canonical basis which is this one:

$$A=\begin{pmatrix} 0 & -1 & -2 \\ 1 & 3 & 1 \\ 1 & 0 & 3 \\ \end{pmatrix}$$ After that i found the characteristic polynomial which is: $(\lambda-2)^3=0$ so we have this polynomial that has only one root with multiplicity 3.

After finding the eigenvalue I found the eigenvector associated to 2, which is $V_3=(-1,0,1)$.

Now the Jordan canonical form should be this one(If I have done it correctly):

$$J=\begin{pmatrix} 2 & 0 & 0 \\ 1 & 2 & 0 \\ 0 & 1 & 2 \\ \end{pmatrix}$$

We know the a Jordan basis is formed with these vectors: $B=v_1,v_2,v_3$

First I found $v_2$ :

$A.v_2=2.v_2+1.v_3$, which is equal to : \begin{align*} \begin{cases} 2x+y+2z &=1\\ x+y+z&=0\\ x+z&=1 \end{cases} \end{align*}So $v_2=(-1,-1,1)$

The problem is here with the last vector: After doing the same operation I get to this point: $A.v_1=2.v_1+1.v_2$, which is equal to: \begin{align*} \begin{cases} 2x+y+2z &=1\\ x+y+z&=0\\ x+z&=1 \end{cases} \end{align*} and clearly this system does not have solution... what am I doing wrong?

• Clearly? The last system not only has a solution, but it has an infinite number of them. – amd Apr 10 '18 at 5:47
• @TheNicouU Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… – user Apr 14 '18 at 14:02

Usually the Jordan normal form is as follow

$$J=\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \\ \end{pmatrix}$$

By Jordan theorem we know that a matrix $P$ exists such that $$P^{-1}AP=J$$

let $$P=[v_1,v_2,v_3,v_4]$$

then P has to satisfy the following system: $$AP=PJ$$ that is in this case $$Av_1=2v_1\implies (A-2I)v_1=0$$ $$Av_2=v_1+2v_2\implies (A-2I)v_2=v_1$$ $$Av_3=v_2+2v_3\implies (A-2I)v_3=v_2$$

Once we have $v_1$ we can find $v_2$ and finally $v_3$, that is

$$(A-2I)v_1=0 \implies\begin{pmatrix} -2 & -1 & -2 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \\ \end{pmatrix}v_1=0 \implies v_1=(1,0,-1)$$

$$(A-2I)v_2=v_1 \implies\begin{pmatrix} -2 & -1 & -2 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \\ \end{pmatrix}v_2=v_1 \implies v_2=(-1,1,0)$$

$$(A-2I)v_3=v_2 \implies\begin{pmatrix} -2 & -1 & -2 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \\ \end{pmatrix}v_3=v_2 \implies v_3=(0,1,0)$$

• in the U.S. it is always taught this way, as far as I know. Evidently there are countries where it is switched. I know France has long done matrices differently, perhaps they still do. – Will Jagy Apr 9 '18 at 23:38
• @WillJagy Yes of course it should be completely equivalent. On wikipedia the upper standard is used en.wikipedia.org/wiki/Jordan_normal_form – user Apr 9 '18 at 23:43
• Here in Uruguay we are taught to do it with 1's below the diagonal. Sorry i thought it would not be a problem for you! Anyway i found my mistake, $v_2=(0,-1,1)$ not $(-1,-1,1)$ – TheNicouU Apr 9 '18 at 23:44
• @TheNicouU Yes I think this is not relevamt at all. Well done then! Bye – user Apr 9 '18 at 23:47

Well, $(A - 2 I )^3 = 0,$ but $(A - 2 I )^2 \neq 0.$ So the minimal polynomial and the characteristic polynomial agree, meaning each eigenvalue occurs in just a single block. Indeed $$(A - 2 I )^2 = \left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 0 & 0 \\ -1 & -1 & -1 \end{array} \right)$$

Now find some $u$ such that $(A - 2 I )^2 u \neq 0.$ Since you want the extra $1$s below the main diagonal, we will put $u$ as the left hand column of $P,$ where we are solving $P^{-1} A P = J$ is the Jordan form you want. I will take $$u = \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right)$$ The middle column will be $v = (A - 2 I) u,$ or $$v = \left( \begin{array}{c} -2 \\ 1 \\ 1 \end{array} \right)$$ and finally $w = (A - 2 I) v =(A - 2 I)^2 u ,$ so that $(A - 2 I) w =(A - 2 I)^3 u = 0 u = 0,$ so that $w$ is a genuine eigenvector $$w = \left( \begin{array}{c} 1 \\ 0 \\ -1 \end{array} \right)$$ The columns of $P$ will be $u,v,w$ so $$P = \left( \begin{array}{ccc} 1 & -2 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{array} \right)$$ next $$P^{-1} = \left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{array} \right)$$

With $$A = \left( \begin{array}{ccc} 0 & -1 & -2 \\ 1 & 3 & 1 \\ 1 & 0 & 3 \end{array} \right)$$ we get to $$\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{ccc} 0 & -1 & -2 \\ 1 & 3 & 1 \\ 1 & 0 & 3 \end{array} \right) \left( \begin{array}{ccc} 1 & -2 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{array} \right) = \left( \begin{array}{ccc} 2 & 0 & 0 \\ 1 & 2 & 0 \\ 0 & 1 & 2 \end{array} \right)$$ $$\left( \begin{array}{ccc} 1 & -2 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{ccc} 2 & 0 & 0 \\ 1 & 2 & 0 \\ 0 & 1 & 2 \end{array} \right) \left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{array} \right) = \left( \begin{array}{ccc} 0 & -1 & -2 \\ 1 & 3 & 1 \\ 1 & 0 & 3 \end{array} \right)$$

If you would like to get the $1$s above the main diagonal instead, start over with the columns of $P$ in order $w,v,u,$ then calculate the new $P^{-1}$ Indeed, the fact that a single Jordan block is similar to its transpose gives a cheap proof that any matrix is similar to its transpose.