I have to calulate A^10.

I have been given A =

(1 2

2 1)

and B =

(1 1

1 -1)

I have calculated that B^-1AB =

(3 0

0 1)

and I don't really understand what my notes mean to take it on from here. Help would be appreciated. :)

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- Feb 8th 2010, 12:25 PMJQ2009Matrix to the Power 10
I have to calulate A^10.

I have been given A =

(1 2

2 1)

and B =

(1 1

1 -1)

I have calculated that B^-1AB =

(3 0

0 1)

and I don't really understand what my notes mean to take it on from here. Help would be appreciated. :) - Feb 8th 2010, 12:43 PMpickslides
To find $\displaystyle A^{10}$

You must find a diagonal matrix $\displaystyle D$ such that

$\displaystyle A^n= BD^nB^{-1} \implies D^n= B^{-1}A^nB$

In your case $\displaystyle n=10$

and where $\displaystyle B$ is the eigenvectors of $\displaystyle A$

So first thing to do is find the eigenvalues and eigenvectors of $\displaystyle A$

(Nod)

It looks like you have been given that (Rofl) so use

$\displaystyle A^n= BD^nB^{-1} $

As you have $\displaystyle D , B$ and $\displaystyle B^{-1}$ - Feb 8th 2010, 12:53 PMpickslides
If that was a little to hard to follow then,

$\displaystyle

\left( {\begin{array}{*{20}{c}}

1& 2 \\

2 & 1\\

\end{array}} \right) ^{10}

=

\left( {\begin{array}{*{20}{c}}

1 & 1 \\

{-1} & 1 \\

\end{array}} \right)

\left( {\begin{array}{*{20}{c}}

3 & 0 \\

0 & 1 \\

\end{array}} \right) ^{10}

\left( {\begin{array}{*{20}{c}}

1 & 1 \\

{-1} & 1 \\

\end{array}} \right) ^{-1}

$

Just solve the RHS.