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Lin alg final coming up, getting the study sesh in

Fun fact:just learned latex so yay

The full problem comes in two parts. Part A is simple/

Given $A$= \begin{pmatrix} 2 & -3/2 \\ 1 & -1/2 \\ \end{pmatrix}

and

$B$= \begin{pmatrix} 3 & -3 \\ 2 & -2 \\ \end{pmatrix}

a. Find a matrix $Q$ such that $Q^{-1}AQ= D$ where $D$ is a diagonal matrix.

Which is simple enough, found the eigen-vectors that came out to be $Q$: \begin{pmatrix} 1 & 3/2 \\ 1 & 1 \\ \end{pmatrix} (which could very well be wrong).

Part B is where the difficulty arises:

b. Show that $\lim_{n\rightarrow\infty} A ^{n} = B$

I'm assuming this has to be solved using some eigen-vectors or some orthogonal basis transformation, however I cannot seem to wrap my head around this problem.

Thank you in advance!

PS. Thanks to John Moeller for the formatting tip.

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  • $\begingroup$ What is $D{}{}$? $\endgroup$ – Euler....IS_ALIVE May 6 '14 at 1:09
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    $\begingroup$ Are $A$ and $B$ supposed to be equal to each other? $\endgroup$ – Braindead May 6 '14 at 1:09
  • $\begingroup$ Re: formatting: you want pmatrix, not array. $\endgroup$ – John Moeller May 6 '14 at 1:14
  • $\begingroup$ Thanks, Ill change that now. I updated the question. $\endgroup$ – Aric May 6 '14 at 1:15
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    $\begingroup$ What's confusing everyone is that $A$ is in fact equal to $B$. $\endgroup$ – John Moeller May 6 '14 at 1:16
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If $A = Q^{-1}DQ$, then $$ A^n = \underbrace{(Q^{-1}DQ)(Q^{-1}DQ) \ldots (Q^{-1}DQ)}_{n\text{ times}} = Q^{-1}D^nQ \text{,} $$ and if $D = \textrm{diag}(d_1,\ldots,d_k)$ then $D^k = \textrm{diag }(d_1^n,\ldots,d_k^n)$.

It follows that for $A = Q^{-1}DQ$ and $D = \textrm{diag }(d_1,\ldots,d_k)$ $$\begin{eqnarray} (1) && \lim_{n\to\infty} A^n \text{ exists if } d_i \in (-1,1] \text{ for all } i, &\text{ and in this case} \\ (2) && \lim_{n\to\infty} A^n = Q^{-1}\hat{D}Q \quad \text{where} & \hat D = \textrm{diag }(\hat d_1,\ldots,\hat d_k),\, \\ &&& \hat d_i = \begin{cases} 1 &\text{if $d_i=1$} \\ 0 &\text{if $-1 < d_i < 1$.} \end{cases} \end{eqnarray}$$

Note that the conditions on the $d_i$ simply guarantee that $\lim_{n\to\infty} d_i$ exists, and $\hat d_i$ is then defined so that $\hat d_i = \lim_{n\to\infty} d_i$. Also note that the $d_i$ are exactly the eigenvalues of $A$.

It follows that, at least for a diagonalizable $A$, $\lim_{n\to\infty} A^n$ exists exactly if all the eigenvalues of $A$ lie within $(-1,1]$.

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  • $\begingroup$ I wish I thought of this. Thanks, I focused too much on the eigenvalues and completely missed this. $\endgroup$ – Aric May 6 '14 at 1:23
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    $\begingroup$ @Klick Well, you need the eigenvalues of $A$, because those are exactly the $d_i$. $\endgroup$ – fgp May 6 '14 at 1:25
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Hint: Since you know that $Q$ is invertible, you know that you can set $A = QDQ^{-1}$, right? So what's $A^2$? $A^3$? $A^n$?

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The matrix $A$ will have eigenvalues $1$ and $\frac{1}{2}$, so $$A^n=Q\pmatrix{1&0\cr0&({\textstyle\frac{1}{2}})^n\cr}Q^{-1}\ .$$ Can you see what you get for the bit in the middle as $n\to\infty$? If you can do this, then since you have already found $Q$, the problem is solved.

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