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I am trying to solve this question

$D$ is the matrix given by $$D = \begin{pmatrix} h & 0 \\ 0 & k \end{pmatrix},$$ where $h = 36$ and $k = -3$. What is the lower-right entry in the matrix $D^n$, where $n = 6$?

This is my attempt, but I am stuck half-way as it just doesn't seem right

enter image description here

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So rather than trying to compute a few powers, you filled in random numbers and began to compute the eigenvalues? Strange... –  rschwieb Oct 21 '12 at 13:04
    
haha my textbook tought me that way..maths.lse.ac.uk/Personal/martin/fme4a.pdf –  JackyBoi Oct 21 '12 at 13:05
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Oh, well that clears things up a bit: you were trying to diagonalize a diagonal matrix. Also strange... The usefulness of diagonalization here is that if you can diagonalize the matrix, then the powers of the matrix are easy to compute, because diagonal matrices are easy to exponentiate. Anyhow, the lesson to be learned here is to definitely spend less time randomly applying things you see in the text, and more time looking at the problem itself. –  rschwieb Oct 21 '12 at 13:06
    
There's also an error in the computation of the eigenvalues: $(-3 - \lambda) = 0$ if and only if $\lambda = -3$ and not $\lambda = 3$. –  TMM Oct 21 '12 at 13:18
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3 Answers

up vote 3 down vote accepted

What do you need the determinant for? It's way easier: check (by induction, for example) that

$$\begin{pmatrix}h&0\\0&k\end{pmatrix}^n=\begin{pmatrix}h^n&0\\0&k^n\end{pmatrix}$$

no matter what $\,h,k,n\,$ are

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clearly this is way easier to do! –  JackyBoi Oct 21 '12 at 13:03
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For a diagonal matrix, the entries of $D^n$ are precisely the $n$th powers of the entries of $D$.

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And eigenvalues of a diagonal matrix are precisely diagonal elements by the definition.

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How does this give the $n^{\text{th}}$ power of $D$? –  robjohn Oct 21 '12 at 13:59
    
Well it does not, but in my opinion it should be usefull to know for the OP, given that he started computing eigenvalues of the diagonal matrix. Anyway having better understanding shall not harm. Don't you think so? –  arkadiy Oct 21 '12 at 18:26
    
Then this might better be a comment to the question. –  robjohn Oct 21 '12 at 20:53
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