# Another question about large powers of matrices

• Aug 17th 2009, 05:45 PM
chella182
Another question about large powers of matrices
So the matrix given is...

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

...and I've established the eigenvectors are $\displaystyle \left(\begin{array}{c}1 \\ 1\end{array}\right)$ and $\displaystyle \left(\begin{array}{c}1 \\ -1\end{array}\right)$ with corresponding eigenvalues $\displaystyle 1$ and $\displaystyle 1-2p$. I'm then asked to show...

$\displaystyle A^n\left(\begin{array}{c}1 \\ 0\end{array}\right)=\frac{1}{2}\left(\begin{array} {c}1 \\ 1\end{array}\right)+\frac{1}{2}(1-2p)^n\left(\begin{array}{c}1 \\ -1\end{array}\right)$

...and I just can't seem to get it, the method I was taught seems to suggest I need to replace the $\displaystyle \left(\begin{array}{c}1 \\ 0\end{array}\right)$ with the eignevalues and eigenvectors multiplied togeter and then added together, but I still can't quite seem to get that expression.
• Aug 17th 2009, 06:04 PM
ynj
Quote:

Originally Posted by chella182
So the matrix given is...

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

...and I've established the eigenvectors are $\displaystyle \left(\begin{array}{c}1 \\ 1\end{array}\right)$ and $\displaystyle \left(\begin{array}{c}1 \\ -1\end{array}\right)$ with corresponding eigenvalues $\displaystyle 1$ and $\displaystyle 1-2p$. I'm then asked to show...

$\displaystyle A^n\left(\begin{array}{c}1 \\ 0\end{array}\right)=\frac{1}{2}\left(\begin{array} {c}1 \\ 1\end{array}\right)+\frac{1}{2}(1-2p)^n\left(\begin{array}{c}1 \\ -1\end{array}\right)$

...and I just can't seem to get it, the method I was taught seems to suggest I need to replace the $\displaystyle \left(\begin{array}{c}1 \\ 0\end{array}\right)$ with the eignevalues and eigenvectors multiplied togeter and then added together, but I still can't quite seem to get that expression.

$\displaystyle A^nX=\lambda^n X,A^n\left(\begin{array}{c}1 \\ 0\end{array}\right)=\frac{1}{2}(A^n\left(\begin{ar ray}{c}1 \\ 1\end{array}\right)+A^n\left(\begin{array}{c}1 \\ -1\end{array}\right))=\frac{1}{2}\lambda_1^n \left(\begin{array}{c}1 \\ 1\end{array}\right)+\frac{1}{2}\lambda_2^n\left(\b egin{array}{c}1 \\ -1\end{array}\right)$
• Aug 17th 2009, 11:06 PM
chella182
Cheers, still not entirely sure why it's $\displaystyle \frac{1}{2}$, but I'm not sure that matters. Although I don't think the $\displaystyle \frac{1}{2}$ was used in the other example I have... oh well.
• Aug 17th 2009, 11:07 PM
ynj
Quote:

Originally Posted by chella182
Cheers, still not entirely sure why it's $\displaystyle \frac{1}{2}$, but I'm not sure that matters. Although I don't think the $\displaystyle \frac{1}{2}$ was used in the other example I have... oh well.

(1,0)=0.5((1,1)+(1,-1))
why not sure?
• Aug 18th 2009, 05:05 AM
chella182
Because that's just repeating what was said and not explaining it, but s'ok, I figured it out in the exam anyway.
• Aug 19th 2009, 04:52 PM
HallsofIvy
This makes two threads in a row in which you have asked for help, got a response which you did not understand and then insulted a person who tried to help you. Do us all a favor and stop posting here.
• Aug 19th 2009, 05:43 PM
ynj
Quote:

Originally Posted by chella182
Because that's just repeating what was said and not explaining it, but s'ok, I figured it out in the exam anyway.

well..may be it is not clear what i have said..sorry..
$\displaystyle A^nX=A^{n-1}AX=A^{n-1}\lambda X=\lambda A^{n-2}AX=\lambda^2 A^{n-2}X=...=\lambda^n X$
then you can substitude all the $\displaystyle \lambda$ to the value given in the exercise.