# Thread: Matrix - Jordan Normal/Canonical Form

1. ## Matrix - Jordan Normal/Canonical Form

QUESTION:

Find $\displaystyle A^k$ when:

A = $\displaystyle \begin{bmatrix} -1 & -2 \\ 2 & 3 \end{bmatrix}$

__________________________________________________ ___

The eigenvalue (of multiplicity 2) is $\displaystyle x = 1$

I've tried putting A into Jordan-Normal/Canonical form, but I find that the matrix $\displaystyle (A- x I)^2 = 0$, and I don't know what to do.

2. Hello, WWTL@WHL!

Find $\displaystyle A^k$ when: .$\displaystyle A \:= \: \begin{bmatrix} \text{-}1 & \text{-}2 \\ 2 & 3 \end{bmatrix}$

I see no choice but to crank out a few powers and seek a pattern . . .

. . $\displaystyle A \;=\;\begin{bmatrix}\text{-}1&\text{-}2\\2&3\end{bmatrix}$

. .$\displaystyle A^2 \;=\;\begin{bmatrix}\text{-}3 & \text{-}4 \\ 4 & 5 \end{bmatrix}$

. .$\displaystyle A^3 \;=\;\begin{bmatrix}\text{-}5&\text{-}6\\6&7\end{bmatrix}$

. .$\displaystyle A^4 \;=\;\begin{bmatrix}\text{-}7&\text{-}8\\8&9\end{bmatrix}$

. .$\displaystyle A^5 \;=\;\begin{bmatrix}\text{-}9&\text{-}10\\10&11\end{bmatrix}$

I think we've found the pattern . . .

We have: .$\displaystyle A^k \:=\:\begin{bmatrix}a&b\\c&d\end{bmatrix}$

. . where: $\displaystyle a$ is the negative of the $\displaystyle k^{th}$ odd number: .$\displaystyle -2k+1$

. . . . . . . $\displaystyle b$ is one less than $\displaystyle a\!:\;\;-2k$

. . . . . . . $\displaystyle c$ is the positive value of $\displaystyle b\!:\;\;2k$

. . . . . . . $\displaystyle d$ is one more than $\displaystyle c\!:\;\;2k+1$

Therefore: . $\displaystyle A^k \;=\;\begin{bmatrix}\text{-}2k+1 & \text{-}2k \\ 2k & 2k+1 \end{bmatrix}$

3. here is a more general way.

$\displaystyle det \begin{bmatrix} -1-x & \text{-}2 \\ 2 & 3-x \end{bmatrix} =(-1-x)(3-x)+4=x^2-2x+1=(x-1)^2$
so we need to find the canonical form. we can see that the matrix has only one eigenvector for 1, so the jordan form is
$\displaystyle J=P^{-1}AP=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$
to find P we can write P=[u,v] where u,v are vectors of size two and
$\displaystyle AP=PJ \rightarrow Av=v, Au=v+u$
so v is an eigenvector, for example (1,-1), and to find u solve the equation $\displaystyle (A-I)u=v$.
now you have $\displaystyle A^k = (PJP^{-1})^k=PJ^k P^{-1}$
notice that this is a general way to find powers of matrices that have the jordan form. because this is used a lot, it is good to know what happens when you raise a jordan matrix to the power of k. to make it easier, write $\displaystyle J=dI+Z$ where the d is the eigenvector (in this case 1) and the Z matrix is a zero matrix with ones only above the main diagonal. in this case Z would just be $\displaystyle \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and then you have
$\displaystyle J^k = (dI+Z)^k = \sum \binom{k}{i} d^{k-i}Z^i$
and the thing about raising Z to the power of i is just just moving the "ones" i times to the top right corner, and here, because Z is only 2x2 matrix, moving the "ones" one time will take them out of the matrix, meaning $\displaystyle Z^2=0$ and because d=1 you get $\displaystyle J^k=\binom{k}{0}I+\binom{k}{1}Z=\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$
this method works on the more general cases. for example, if you have more than one jordan block, then you can do this for every block separately.

4. Originally Posted by WWTL@WHL
QUESTION:

Find $\displaystyle A^k$ when:

A = $\displaystyle \begin{bmatrix} -1 & -2 \\ 2 & 3 \end{bmatrix}$

__________________________________________________ ___

The eigenvalue (of multiplicity 2) is $\displaystyle x = 1$

I've tried putting A into Jordan-Normal/Canonical form, but I find that the matrix $\displaystyle (A- x I)^2 = 0$, and I don't know what to do.

Yes, of course $\displaystyle (A- I)^2= 0$ because every matrix satisfies its own characteristic equation.

First look for a eigenvectors for $\displaystyle \lambda= 1$:
$\displaystyle \begin{bmatrix}-1 & -2 \\ 2 & 3\end{matrix}\begin{matrix} x \\ y\end{matrix}= \begin{matrix} x \\ y\end{bmatrix}$
which gives the equations -x- 2y= x and 2x+ 3y= y. Those both reduce to y= -x so <1, -1> is an eigenvector but we don't find a second. But because $\displaystyle (A- I)^2 v= 0$ for all vectors v and there are some such that (A- I)v is not 0, there must be v such that (A- I)((A-I)v)= 0 which means (A-I)v is an eigenvalue, a multiple of <1, -1>. That is, look for v such that $\displaystyle (A- I)v= \begin{bmatrix}-2 & -2 \\ 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}1 \\ -1\end{bmatrix}$
That gives the equations -2x-2y= 1 and 2x+ 2y= -1 which are, again, equivalent. But taking x= 1 we get y= -3/2 so <1, -3/2> is a "generalized eigenvalue". Create the matrix $\displaystyle \begin{bmatrix}1 & 1 \\ -1 & -3/2\end{bmatrix}$ with those vectors as columns and find its inverse, $\displaystyle \begin{bmatrix}3 & 2 \\ -2 & -2\end{matrix}$

Now we have $\displaystyle \begin{bmatrix}3 & 2 \\ -2 & -2\end{bmatrix}\begin{bmatrix}-1 & -2 \\ 2 & 3\end{bmatrix}\begin{bmatrix}1 & 1 \\ -1 & -3/2\end{bmatrix}= \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$, the "Jordan Normal Form" for A.

That is, $\displaystyle P^{-1}AP= N$ so $\displaystyle A= PNP^{-1}$ and $\displaystyle A^n= PN^nP^{-1}$

And, of course, $\displaystyle N^n= \begin{bmatrix}1 & n \\ 0 1$ so to find $\displaystyle A^n$, you just need to calculate $\displaystyle \begin{bmatrix}1 & 1 \\ -1 & -3/2\end{bmatrix}\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\begin{bmatrix}3 & 2 \\ -2 & -2\end{bmatrix}$.