Matrices help

**chella182** · August 17th 2009, 03:50 PM

Okay, I'll write out the question in full, but my problem is very specific to something that crops up later on in the solution I have here. Sorry if this is in the wrong subforum, but the module's called "Linear Methods", so I thought this seemed appropriate.

Basically it's a method to compute large power of matrices, and the example gives

$\left(\begin{array}{cc}0.9 & 0.2 \\ 0.1 & 0.8\end{array}\right)^n\left(\begin{array}{c}x_1 \\ x_2\end{array}\right)$ or $\underline{\underline{A}}^n\underline{v}$ , as my lecturer likes to write stuff.

The notes then say "It would be easy to compute $\underline{\underline{A}}^n\underline{v}$ if $\underline{v}$ were an eigenvector i.e. $\underline{\underline{A}}^n\underline{v}=\lambda^n \underline{v}$ ". Up to this point, I undersand completely.

Then we start finding the eigenvectors and the calculation goes as such

$\left(\begin{array}{cc}0.9 & 0.2 \\ 0.1 & 0.8\end{array}\right)\left(\begin{array}{c}1 \\ -1\end{array}\right)=\left(\begin{array}{c}0.9\time s1-0.2\times1 \\ 0.1\times1-0.8\times1\end{array}\right)$
$=\left(\begin{array}{c}0.7 \\ -0.7\end{array}\right)=0.7\color{green}\boxed{\left (\begin{array}{c}1 \\ -1\end{array}\right)}$

...and he notes that the green box is always an eigenvector. So far, I still understand. It's this next bit that's confused me. He notes that $1$ is always an eigenvalue, but he just seems to have plucked this vector out of the air (the one in the red box) and I really can't see where it's came from. It's probably very simple but this is my problem with the example/solution.

$\left(\begin{array}{cc}0.9 & 0.2 \\ 0.1 & 0.8\end{array}\right)\color{red}\boxed{\left(\begi n{array}{c}2/3 \\ 1/3\end{array}\right)}\color{black}=1\left(\begin{ar ray}{c}0.9(2/3)+0.2(1/3) \\ 0.1(2/3)+0.8(1/3)\end{array}\right)$

$=1\left(\begin{array}{c}2/3 \\ 1/3\end{array}\right)$

So basically, where has the $\mathbf{\left(\begin{array}{c}2/3 \\ 1/3\end{array}\right)}$ came from? Sorry this is a bit long winded by the way, I just thought having the whole question might help.

**ynj** · August 17th 2009, 05:16 PM

you can solve the equation $AX=X\leftrightarrow (A-I)X=0$ , which is a linear equation groups.

**hatsoff** · August 17th 2009, 05:16 PM

If $\lambda=1$ is an eigenvalue, then $(A-\lambda I_2)v=0$ yields our eigenvectors. Specifically, we have

$\left(\begin{array}{cc}0.9-1 & 0.2 \\ 0.1 & 0.8-1\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)=\left(\begin{array}{cc}-0.1 & 0.2 \\ 0.1 & -0.2\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)=\left(\begin{array}{c}0.2y-0.1x \\ 0.1x-0.2y\end{array}\right)=\left(\begin{array}{c}0 \\ 0\end{array}\right)$ ,

which gives us $0.1x=0.2y$ $\Rightarrow$ $x=2y$ .

So a basis for $\mathsf{E}_1$ is

$\beta=\left\{\left(\begin{array}{c}2 \\ 1\end{array}\right)\right\}$ .

Therefore, we have

$\frac{1}{3}\left(\begin{array}{c}2 \\ 1\end{array}\right)=\left(\begin{array}{c}2/3 \\ 1/3\end{array}\right)\in\mathsf{E}_1$ .

**chella182** · August 17th 2009, 05:24 PM

Okay, I'm with you right up until the $\frac{1}{3}$ comes into it at the end... where did that come from?

**hatsoff** · August 17th 2009, 05:56 PM

Originally Posted by chella182

Okay, I'm with you right up until the $\frac{1}{3}$ comes into it at the end... where did that come from?

Any linear combination of vectors in $\beta$ is going to be in the eigenspace $\mathsf{E}_1=\text{span}(\beta)$ . So you can throw in any value you want, as long as it's in the field, a subset of which appears to be $\mathbb{R}$ .

**chella182** · August 17th 2009, 11:04 PM

I still don't quite understand :\ oh well, exam in an hour and a half.

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