Ok I recalculated and got

[-(4/5)*2^n+ (9/5)*.5^n (6/5)*2^n - (1/5)*.5^n

-(4/5)*3^n- (9/5)*.5^n (6/5)*3^n + (1/5)*.5^n]

for the product of the three matricies

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- Jul 29th 2010, 06:48 PMSpiffyEh
Ok I recalculated and got

[-(4/5)*2^n+ (9/5)*.5^n (6/5)*2^n - (1/5)*.5^n

-(4/5)*3^n- (9/5)*.5^n (6/5)*3^n + (1/5)*.5^n]

for the product of the three matricies - Jul 30th 2010, 02:55 AMAckbeet
Closer, but still no cigar. I think you're making arithmetic errors in your matrix multiplication. You've got

$\displaystyle P=\begin{bmatrix}

2 &-1\\

3 &1

\end{bmatrix},\quad D^{n}=\begin{bmatrix}

1 &0\\

0 &(-0.5)^{n}

\end{bmatrix},\quad\text{and}\quad P^{-1}=\begin{bmatrix}

\frac{1}{5} &\frac{1}{5}\\

-\frac{3}{5} &\frac{2}{5}

\end{bmatrix}.$

Now, one step at a time. Show me either $\displaystyle PD^{n}$ or $\displaystyle D^{n}P^{-1}.$ Then show me $\displaystyle PD^{n}P^{-1}.$ - Jul 30th 2010, 12:22 PMSpiffyEh
$\displaystyle PD^{n}=\begin{bmatrix}

2 &(0.5)^{n}\\

3 &(-0.5)^{n}

\end{bmatrix},\quad\text{and}\quad PD^{n}P^{-1}=\begin{bmatrix}

\frac{2}{5}-\frac{3}{5}(0.5)^{n} &\frac{2}{5}+\frac{2}{5}(0.5)^{n}\\

\frac{3}{5}+\frac{3}{5}(0.5)^{n} &\frac{3}{5}-\frac{2}{5}(0.5)^{n}

\end{bmatrix}.$

I recalculated, is it correct this time? - Jul 30th 2010, 12:53 PMAckbeet
I get

$\displaystyle PD^{n}=

\begin{bmatrix}

2 &-(-0.5)^{n}\\

3 &(-0.5)^{n}

\end{bmatrix},$ and

$\displaystyle PD^{n}P^{-1}=\begin{bmatrix}

0.6(-0.5)^{n}+0.4 &-0.4(-0.5)^{n}+0.4\\

-0.6(-0.5)^{n}+0.6 &0.4(-0.5)^{n}+0.6

\end{bmatrix}.$

You're even closer than before. Just watch those minus signs!

So what happens next? - Jul 30th 2010, 01:09 PMSpiffyEh
Can I trake the limit from here? Or do I multiply by x vector first?

- Jul 30th 2010, 01:12 PMAckbeet
It probably makes no difference; however, to play it safe, multiply by the x vector first.

- Jul 30th 2010, 02:00 PMSpiffyEh
Ok so if I multiply that in I should get a 2x1 matrix correct? Then I take the limit

- Jul 30th 2010, 03:15 PMAckbeet
Correct. What do you get?

- Jul 30th 2010, 08:10 PMSpiffyEh
$\displaystyle PD^{n}P^{-1}X=\begin{bmatrix}

(0.6(-0.5)^{n}+0.4)x_{1} + (-0.4(-0.5)^{n}+0.4)x_{2}\\

(-0.6(-0.5)^{n}+0.6)x_{1} + (0.4(-0.5)^{n}+0.6)x_{2}

\end{bmatrix}.$ - Jul 31st 2010, 03:01 AMAckbeet
Right. But you know that $\displaystyle x_{1}+x_{2}=1$. That turns out to be important for computing the limit. I would eliminate one of those components. What does that give you?

- Jul 31st 2010, 09:39 AMSpiffyEh
umm.. I'm not sure if this is correct but would i distribute the x's and cancel out the [tex]0.4x_{1}+.01x_{2} and the same for the 0.6 one?

- Jul 31st 2010, 10:07 AMAckbeet
Well, you know that $\displaystyle x_{1}+x_{2}=1.$ Therefore, $\displaystyle x_{2}=1-x_{1}.$ Hence, I can represent the $\displaystyle x$ vector as

$\displaystyle x=\begin{bmatrix}x_{1}\\ 1-x_{1}\end{bmatrix}.$

Multiply your $\displaystyle PD^{n}P^{-1}$ times this vector, and then take the limit. What do you get here? - Jul 31st 2010, 10:32 AMSpiffyEh
$\displaystyle PD^{n}P^{-1}X=\begin{bmatrix}

(0.6(-0.5)^{n}+0.4)x_{1} + (1-x_{1})(-0.4(-0.5)^{n}+0.4)\\

(-0.6(-0.5)^{n}+0.6)x_{1} + (1-x_{1})(0.4(-0.5)^{n}+0.6)

\end{bmatrix}.$

taking the limit, wouldn't all the ^n terms go to 0? So then I have

$\displaystyle \begin{bmatrix}

(0.4)x_{1} + (1-x_{1})(0.4)\\

(0.6)x_{1} + (1-x_{1})(0.6)

\end{bmatrix}.$

so the terms cancel to

$\displaystyle \begin{bmatrix}

0.4\\

0.6

\end{bmatrix}.$ - Jul 31st 2010, 10:38 AMAckbeet
Excellent! In your write-up, you might want to mention why all the exponentiated terms vanish in the limit.

So, that pretty much wraps up your first two problems. Have you gotten a good start on the third problem?

Just as a heads-up: it is considered better both to write up your original problem in LaTeX, as well as not to include more than two related problems in one thread.

I enjoyed these problems myself. Have a good one! - Jul 31st 2010, 11:02 AMSpiffyEh
Thank you so much for guiding me through it, it helps so much more than just giving me an answer.

I'm trying to do the third but I can't find a relationship between eigenvectors of A and A transpose. The eigenvalues are the same but that doesn't mean it will have all the eigenvectors unless its guaranteed you get an eigenvector when you have an eigenvalue, which i'm pretty sure it's not.

Thanks for the heads-up, I'll take the time to write it up next time, it just takes a while since I'm not too familiar with the LaTeX