how can you determine patterns of matrices to the nth power?

Question

i was given the matrices: P=[{3,1}{1,3}] and S=[{4,2}{2,4}].
the problem then said: P^2=[{2,1}{1,3}]^2=[{10,6}{6,10}]=2[{5,3}{3,5}]; and S^2=[{4,2}{2,4}]^2=[{20,16}{16,20}]=2[{10,8}{8,10}]. Calculate P^n and S^n for other values of n and describe an patterns you observe. Can anyone help? i can't find any patterns. thanks!!!

Answer 1

Hi.. This one will require few steps for explaining, so pls bear with a long answer.

First, lets call any such matrix as {(a,b),(b,a)} to generalize the numbers. Then, every power n of such a matrix will always have a form, {(X,Y),(Y,X)}, where the values of X,Y will be different for each power, and also show a pattern that I will explain in a moment.

Also, X+Y = (a+b)^n will always hold. This means that in case of the example you have given, any power of P will be of the form {(X,Y),(Y,X)}, and X+Y will be a power of 4 (as your a,b are 3,1).

Secondly, the element X will be a^n, and all those terms of the binomial expansion of (a+b)^n, which are contain the element a raised to the power n, n-2, n-4, n-6 etc. till either a^1, or a^0 depending on n being odd, or even.

Similarly, the element Y will contain all the alternating powers of a, commencing from n-1, till power 1 or 0 depending on n being odd or even.

So, for any n, the term X is a^n+C(n,2)a^(n-2)b^2+C(n,4)a^(n-4)b^4+C(n,6)a^(n-6)b^6.. and the term Y is C(n,1)a^(n-1)b+C(n,3)a^(n-3)b^3...

or in simple English, the terms are sums of alternating terms of Binomial expansion of (a+b)^n, expressed as {(X,Y),(Y,X)}.

An example is say, p={(a,b),(b,a)}. Then p^5 will be of the form {(X,Y),(Y,X)}, where,

X= a^5+C(5,2)a^3b^2+C(5,4)ab^4 = a^5+10a^3b^2+5ab^4
Y= C(5,1)a^4b+C(5,3)a^2b^3+b^5 = 5a^4b+10a^2b^3+b^5

If we take specific example of P={(3,1),(1,3)}, then,
P^5= {(528,496),(496,528)}, as, 3^5+10.3^3+5.3=528 and 5.3^4+10.3^2+1 = 496

trust this helped :-)

Answer 2

The only pattern I observe is that the two diagonals are always contain equal elements, as do the original matrices:

P^2 = [{10,6}{6,10}] P^3 = [{36,28}{28,36}] p^4 = [{136,120}{120,136}]

S^2 = [{20,16}{16,20}] S^3 = [{112,104}{104,112}] S^4 = [{656,640}{640,656}]

Answer 3

you can easily find the nth power of a matrix if the matrix is diagonalizable. A matrix A is diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix P such that
(P-1)AP =D, where (P-1) is inverse of P and D is a diagonal matrix.

In your case, you are given with symmetric matrices, you can always find such a matrix P which will also be orthogonal. i.e., inverse of P = transpose of P.

To find the matrix P and D , you have to find eigenvalues and eigenvectors of A.
I am solving one for you.

A= [{4,2}{2,4}]. The eigenvalues of A are 2 and 6. the eigenvector corresponding to 2 is[1,-1]. the eigenvector corresponding to 6 is[1,1]. Then diagonal matrix will be diag{2,6}. P will be formed by eigenvectors of 2 and 6 as a column vectors of P in the same order as eigenvalues 2,6 are presented in the diagonal matrix D.

Hence, A = PD(P-1), and A^n = P(D^n) (P-1). In this case,
[{4,2}{2,4}] = (1/2) [{1,1}{-1,1}] [{2,0}{0,6}] [{1,-1}{1,1}].
and
[{4,2}{2,4}] ^n = (1/2)[{1,1}{-1,1}] [{2,0}{0,6}]^n [{1,-1}{1,1}].
= (1/2)[{1,1}{-1,1}] [{2^n,0}{0,6^n}] [{1,-1}{1,1}].
..........................................(*)
and [{4,2}{2,4}] ^2 = (1/2)[{1,1}{-1,1}] [{2^2,0}{0,6^2}] [{1,-1}{1,1}].
= (1/2)[{1,1}{-1,1}] [{4,0}{0,36}] [{1,-1}{1,1}].
= [{20,16}{16,20}].

You can find other powers using the formula (*).

Answer 4

In general terms, and for any square matrix, there is a very useful shortcut for taking matrix powers, called the Cayley-Hamilton Theorem. It is described in the link below:

Answer 5

I will give a hint: Notice that the matrices are diagonalizeable.

Answer 6

No matter how many steps that simple guy tells you it takes, just do it he knows what he is talking about :-)