# Matrix - Jordan Normal/Canonical Form

• Feb 28th 2009, 06:19 PM
WWTL@WHL
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.

• Feb 28th 2009, 07:03 PM
Soroban
Hello, WWTL@WHL!

Quote:

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

• Mar 1st 2009, 10:18 PM
Prometheus
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.
• Mar 2nd 2009, 06:33 AM
HallsofIvy
Quote:

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 $\displaystyle \begin{bmatrix}1 & 1 \\ -1 & -3/2\end{bmatrix}\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\begin{bmatrix}3 & 2 \\ -2 & -2\end{bmatrix}$.