I'm rather unsure what is going on with the last part of question c any help would be greatly appreciated!

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- Mar 30th 2011, 10:48 AMLHSDiagonalizable matrix used in polynomial form
I'm rather unsure what is going on with the last part of question c any help would be greatly appreciated!

http://img163.imageshack.us/img163/4992/photo1mpl.jpg - Mar 30th 2011, 10:59 AMFernandoRevilla
- Mar 30th 2011, 11:12 AMLHS
sorry.. was suffering from some technical difficulties!

- Mar 31st 2011, 02:12 PMmasnarski
Hi there,

Your matrix is diagonalizable. In particular,

$\displaystyle

$B=PDP^{-1}$, where

$D=\begin{pmatrix}

6 & 0 & 0 \\

0 & 6 & 0 \\

0 & 0 & 3\\

\end{pmatrix}\\

$

Hence,

$\displaystyle

B^n = (PDP^{-1})^n = (PDP^{-1})(PDP^{-1}) ... (PDP^{-1})\\

= PD^nP^{-1}\\

$

Substituting this into your equation yields,

$\displaystyle

a_nPD^nP^{-1} + ... + a_1PDP^{-1} + a_0& = C\\

$

and pre and post multiplying both sides by P inv. and P (respectively) yields:

$\displaystyle

a_nD^n + ... + a_0& = P^{-1}CP\\

$

where C is that diagonal matrix you are given on the right-hand side of the equation (with entries 1, 2, 3).

However, you will notice that when you multiply P^{-1}CP out, you will get a matrix that is not diagonal on the right hand side of the last equation.

But a power of a diagonal matrix is again a diagonal matrix, a scalar multiple of a diagonal matrix is also diagonal, and the sum of diagonal matrices is again diagonal. Hence, you will never get the RHS, so it is impossible. - Mar 31st 2011, 02:25 PMLHS
Can't really thank you enough.. looks worryingly simple now! You know, one of those questions you have a mental block on, that was bothering the life out of me

- Mar 31st 2011, 07:02 PMmasnarski
Don't worry about it. It's something I might not have figured out some other day.

When I am facing problems like this, and a potential way of solving it doesn't occur to me within a few minutes, I consciously remind myself of "what I have in my toolbox." This involves not only summoning up what I know about diagonalizable matrices (and perhaps even making lists and drawing arrows to indicate potential connections between ideas), but also reminding myself of the basic methods of proof: direct proof, contrapositive, contradiction, etc. When in doubt -- proof by induction! - Apr 1st 2011, 01:54 AMLHS
That's good advice, i'll bear it in mind :) cheers mate!