Calculating powers of matrices

[rdajunior95]

Hi,


I get a lot of questions about calculating M^k, where M is a square matrix!

They say you can use an equation like M^k=PD(P^-1) where D is a diagonal matrix.

I don't know how to calculate this!

Any help will be appreciated!


P.S. Sorry if this is in the wrong section!

 

[D H]

Who are "they"?

This sounds a bit too much like a homework question, so for now I will not give a complete answer. Some square matrices can be decomposed such that M=PDP^-1. These are the diagonalizable matrices.

If some matrix M is diagonalizable, what is M²? Try writing it out. Remember that while matrix multiplication is not commutative, it is associative.

 

[rdajunior95]

they are the people who set out the question paper of 9231 Further maths.

So m^2 will be (PD(P^-1))*(PD(P^-1))

so P * P^-1 = I so it will be PDPD.

Am I right?

 

[rdajunior95]

Also I have a question that find a non-singular matrix P and a diagonal matrix D such that (A^5) = PD(P^-1)

Where A is a 3x3 matrix

A is given but I don't know how to put a matrix in a post!

Sorry

 

[D H]

they are the people who set out the question paper of 9231

What does this mean? I realize that while English may not be your native language, this cryptic statement wouldn't make sense in any language. What is the "question paper of 9231"?

Further maths.

So m^2 will be (PD(P^-1))*(PD(P^-1))

so P * P^-1 = I so it will be PDPD.

Am I right?

No. Try again.

 

[rdajunior95]

I think you have misread. What I have typed is 9231 Further maths!

And please tell me from where to start!

I haven't even read about it in any books of mine.

I have Complete series of Further Maths 1, 2 & 3 for OCR and Further Pure Mathematics by brian gaulter and mark gaulter and complete series by L. Bostock

 

[Mark44]

You need to give the context that 9231 Further Maths is a course or placement test at Cambridge International Education (CIE), something like that. The only reason I know this is that you provided a link to the syllabus for 2010 in another thread. Without this information, people will have no idea what you're talking about.

 

[HallsofIvy]

An n by n matrix is "diagonalizable" if and only it has n independent eigenvectors. (That is always true if A is a symmetric matrix and/or has n distinct eigenvalues.)

If A has n independent eigenvectors, the form the matrix P having those eigenvectors as eigenvalues.

Then \(\displaystyle P^{-1}AP= D\) is a diagonal matrix having the eigenvalues of A on the diagonal.

It is easy to take the nth power of a diagonal matrix- \(\displaystyle D^n\) is the diagonal matrix having the nth powers of its diagonal elements on the diagonal.

Also, if $D= P^{-1}AP$, then $D^n= (P^{-1}AP)(p^{-1}AP)\cdot cdot\cdot(P^{-1}AP)= P^{-1}A^nP$ since all those internal "P"s and "$P^{-1}$"s cancel.

So $A^n= PD^nP^{-1}$.

Most matrices are NOT "diagonalizable" but can be reduced to "Jordan normal form" which has some "1"s above the diagonal. The formula for nth powers is more complicated but still doable.

 

[rdajunior95]

Thanks for the help! :D