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

What is $\displaystyle A^{1000}$

It's not diagonalizable.

Printable View

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

What is $\displaystyle A^{1000}$

It's not diagonalizable. - Apr 13th 2009, 12:04 PMMush
- Apr 13th 2009, 12:14 PMNonCommAlg
one way is to use induction to show that $\displaystyle A^n=\begin{pmatrix}1 & -2n \\ 0 & 1 \end{pmatrix}.$ a better way, which can be used for more complicated matrices, is to write $\displaystyle A=I + B,$ where $\displaystyle I$ is the identity matrix and

$\displaystyle B=\begin{pmatrix}0& -2 \\ 0 & 0\end{pmatrix}.$ then $\displaystyle B^2=0$ and $\displaystyle I,B$ commute. then by the binomial theorem: $\displaystyle A^n=(I+B)^n=I^n + nB=I + nB=\begin{pmatrix}1 & -2n \\ 0 & 1 \end{pmatrix}.$ - Apr 13th 2009, 12:35 PMSpec
- Apr 13th 2009, 12:55 PMNonCommAlg
this is a good method of finding powers of matrices. as another example we want to find $\displaystyle A^n$ where $\displaystyle A=\begin{pmatrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}.$ so we let $\displaystyle B=\begin{pmatrix}0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix}.$

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

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