# [SOLVED] Matrix^1000

• Apr 13th 2009, 11:58 AM
Spec
[SOLVED] Matrix^1000
$A=\left(\begin{array}{cc}1&-2\\0&1\end{array}\right)$

What is $A^{1000}$

It's not diagonalizable.
• Apr 13th 2009, 12:04 PM
Mush
Quote:

Originally Posted by Spec
$A=\left(\begin{array}{cc}1&-2\\0&1\end{array}\right)$

What is $A^{1000}$

It's not diagonalizable.

$
\left(\begin{array}{cc}1&-2\\0&1\end{array}\right)^n = \left(\begin{array}{cc}1&-2n\\0&1\end{array}\right)

$
• Apr 13th 2009, 12:14 PM
NonCommAlg
Quote:

Originally Posted by Spec
$A=\left(\begin{array}{cc}1&-2\\0&1\end{array}\right)$

What is $A^{1000}$

It's not diagonalizable.

one way is to use induction to show that $A^n=\begin{pmatrix}1 & -2n \\ 0 & 1 \end{pmatrix}.$ a better way, which can be used for more complicated matrices, is to write $A=I + B,$ where $I$ is the identity matrix and

$B=\begin{pmatrix}0& -2 \\ 0 & 0\end{pmatrix}.$ then $B^2=0$ and $I,B$ commute. then by the binomial theorem: $A^n=(I+B)^n=I^n + nB=I + nB=\begin{pmatrix}1 & -2n \\ 0 & 1 \end{pmatrix}.$
• Apr 13th 2009, 12:35 PM
Spec
Quote:

Originally Posted by NonCommAlg
$...=(I+B)^n=I^n + nB=...$

I don't get that part. I know what the binomial theorem says, but I've never seen it used on matrices.

EDIT: Nevermind, I didn't see the B^2 part.
• Apr 13th 2009, 12:55 PM
NonCommAlg
Quote:

Originally Posted by Spec

I don't get that part. I know what the binomial theorem says, but I've never seen it used on matrices.

EDIT: Nevermind, I didn't see the B^2 part.

this is a good method of finding powers of matrices. as another example we want to find $A^n$ where $A=\begin{pmatrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}.$ so we let $B=\begin{pmatrix}0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix}.$

then $A=I+B, \ B^2=\begin{pmatrix}0 & 0 & ac \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},$ and $B^3=0.$ again $I,B$ commute and so we can use binomial theorem, which gives us:

$A^n=(I+B)^n=I + nB + \frac{n(n-1)}{2}B^2 = \begin{pmatrix}1 & na & nb + \frac{n(n-1)}{2}ac \\ 0 & 1 & nc \\ 0 & 0 & 1 \end{pmatrix}.$