matrix powers

[maximus_koncept7]

hi guys, today i was confonted with this problem in grade 11 high school math.

consider the matrix M=(2 0)
(0 2)

calculate M^n for n = 2,3,4,5,10,20,50

and find a general expression for the matrix M^n in terms of n.

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this problem has troubled me a lot, and no matter how hard i tried, i couldnt find a solution...could someone help me out here?
thanks guys

[shmoe]

Find M^2, M^3, M^4 and M^5 by hand. What do you get? do you see a pattern?

[maximus_koncept7]

yeah...the first and the last element are 2^the power given.

but how do i put this into a equation?
thanks again for all your help!

[artistatwork]

M^n = M*2^(n-1) is the equation for this particular matrix only. The equation differs for others.

[HallsofIvy]

Since you expressed M as
(2 0)
(0 2)
what's wrong with Mn as
(2n 0)
(0 2n) ?

[Ibnerd]

I have the same question and am just wondering what your final general formula was... the (2^n 0)
(0 2^n) one only works when there are zeros on the diagonal and I have a question where the matrix is (3 1)
(1 3)
-I don't understand the outcome answers when it is squared :
(10 6)
(6 10) -is there a general formula to use? What are matrices like this called?

[John Creighto]

I have the same question and am just wondering what your final general formula was... the (2^n 0)
(0 2^n) one only works when there are zeros on the diagonal and I have a question where the matrix is (3 1)
(1 3)
-I don't understand the outcome answers when it is squared :
(10 6)
(6 10) -is there a general formula to use? What are matrices like this called?

Any matrix with unique eigenvectors and eigen values can be diagonalized:

A=V D V^-1

Where D is a diagonal matrix of the eign values
and
The ith column of V is the eign vector which corresponds to the eign value on the diagnal of the ith column of D.

A^2=V D V^-1 V D V^-1=V D (V^-1 V) D V^-1=
V D I D V^-1=V D D V^-1=V D^2 V^-1

In general

A^N=V D^N V^-1

Your above expression is simple enough that just by doing regular multiplication you might be able to see the pattern without applying the above theory.

[Rainbow Child]

The matrix M can be written as M=2\,I_2, where I_2 is the unit 2\times2 matrix. Thus M^2=2\,I_2\cdot 2\,I_2=2^2\, I_2^2=2^2\,I_2. Try the same thing for M^3,\,M^4,\dots If you want to prove that M^n=2^n\,I_2 use induction.

[Rainbow Child]

Just noticed that this thread is almost 4 months old! :rofl:

[Ibnerd]

Thanks for the help-is there a simpler way to prove that-I've never heard of the stuff you are referring to...this is for grade 11 math. Thanks again though!

[HallsofIvy]

Well, I didn't learn it until I was in college!

[masterprimus]

I got the exact same problem, and am pretty sure that there is some relatively 11th grade adequate way to approach this. Maybe something with step 3, where (k+1 k-1) this matrix applies to both the example (2 0) and (3 1)
_________________(k-1 k+1) __________________________________(0 2) ___(1 3)

I really can't find a general equation for the last one in terms of k and n...

[masterprimus]

After some serious work on my last post i came up with this, feedback please!
This is matrix Mk raised to the nth power.

M^{N}_{K} = \left([(k+1) +(k - 1)\sum^{n}_{x=1} k^{x}] [(k-1) +(k - 1)\sum^{n}_{x=1} k^{x}] \right)
\left([(k - 1) +(k - 1)\sum^{n}_{x=1} k^{x}] [(k + 1) +(k - 1)\sum^{n}_{x=1} k^{x}] \right)

(The thumbnail is a lot clearer)