# Thread: linear algebra - concept proofs

1. 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

2. 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 $\displaystyle PD^{n}P^{-1}.$

3. $\displaystyle PD^{n}=\begin{bmatrix} 2 &(0.5)^{n}\\ 3 &(-0.5)^{n} \end{bmatrix},\quad\text{and}\quad PD^{n}P^{-1}=\begin{bmatrix} \frac{2}{5}-\frac{3}{5}(0.5)^{n} &\frac{2}{5}+\frac{2}{5}(0.5)^{n}\\ \frac{3}{5}+\frac{3}{5}(0.5)^{n} &\frac{3}{5}-\frac{2}{5}(0.5)^{n} \end{bmatrix}.$

I recalculated, is it correct this time?

4. 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?

5. Can I trake the limit from here? Or do I multiply by x vector first?

6. It probably makes no difference; however, to play it safe, multiply by the x vector first.

7. Ok so if I multiply that in I should get a 2x1 matrix correct? Then I take the limit

8. Correct. What do you get?

9. $\displaystyle PD^{n}P^{-1}X=\begin{bmatrix} (0.6(-0.5)^{n}+0.4)x_{1} + (-0.4(-0.5)^{n}+0.4)x_{2}\\ (-0.6(-0.5)^{n}+0.6)x_{1} + (0.4(-0.5)^{n}+0.6)x_{2} \end{bmatrix}.$

10. 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?

11. 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?

12. 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?

13. $\displaystyle PD^{n}P^{-1}X=\begin{bmatrix} (0.6(-0.5)^{n}+0.4)x_{1} + (1-x_{1})(-0.4(-0.5)^{n}+0.4)\\ (-0.6(-0.5)^{n}+0.6)x_{1} + (1-x_{1})(0.4(-0.5)^{n}+0.6) \end{bmatrix}.$

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
$\displaystyle \begin{bmatrix} 0.4\\ 0.6 \end{bmatrix}.$

14. 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!

15. 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

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