# Matrix induction

1. Aug 13, 2009

### boneill3

1. The problem statement, all variables and given/known data

Given the following matrix A find An

$A= $\left( \begin{array}{ccc} a & -b \\ b & a \\ \end{array} \right)$$

2. Relevant equations

I am using matrix multiplication and eventually mathematical induction

3. The attempt at a solution

First I find A2

$A^2= $\left( \begin{array}{cc} a & -b \\ b & a \\ \end{array} \right) \times \left( \begin{array}{cc} a & -b \\ b & a \\ \end{array} \right)$$

$A^2= $\left( \begin{array}{cc} a^2-b^2 & -2ab \\ 2ab & a^2-b^2 \\ \end{array} \right)$ $A^3=AA^2 \[ \left( \begin{array}{cc} a & -b \\ b & a \\ \end{array} \right) \times \left( \begin{array}{cc} a^2-b^2 & -2ab \\ 2ab & a^2-B^2 \\ \end{array} \right)$$

$A^3= \[ \left( \begin{array}{cc} a(a^2-3b^2) & -b(3a^2-b^2) \\ b(3a^2-b^2) & a(a^2-3b^2) \\ \end{array} \right)$

$A^4=AA^3 \[ \left( \begin{array}{cc} a & -b \\ b & a \\ \end{array} \right) \times \left( \begin{array}{cc} a(a^2-3b^2) & -b(3a^2-b^2) \\ b(3a^2-b^2) & a(a^2-3b^2) \\ \end{array} \right)$

$A^4= \[ \left( \begin{array}{cc} a^4-6a^2b^2+b^4 & -4a(a^2-b^2)b \\ 4a(a^2-b^2)b & a^4-6a^2b^2+b^4 \\ \end{array} \right)$

now I am trying to look for a patern so that I can use that to find An

I see that The diagonals are the same except that row 1 column 2 is negative.

for row 1 column 1 I think it might be something like

$a^n (n-1+n)a^{n-2}$

and for row 2 column 1 I think it may be

$na^{n-1}b-nab^{n-1}$

As they are the same diagonals I can use the same ones except for row 1 column 2 which I make negative.

If I could someone can see a pattern I would greatly appreciate it as Then I can try to prove it using induction.

regards
Brendan

2. Aug 14, 2009

### kNYsJakE

If a matrix A is given, you probably already had learn that you can set a matrix A as $$A=QDQ^{-1}$$

where Q is an appropriate matrix and D is a diagonal matrix.

Now $$A^{n}=(QDQ^{-1})(QDQ^{-1})\cdots(QDQ^{-1})$$
$$=QD^{n}Q^{-1}$$

It should be pretty straight from this point.

I hope it helped =)

Last edited: Aug 14, 2009
3. Aug 15, 2009

### boneill3

It is a bit advanced for me though as I'm not up too that level yet.
I have just started linear Algebra so we have been doing exercices to show that certain patterns can form when doing matrix powers.
So I am supposed to find the general expression just by recognising a pattern.
Perhaps my calculations are wrong.
regards
Brendan

4. Aug 15, 2009

### boneill3

I was looking through some class notes examples. I've come up with a pattern which is close but but the sign of the expression is wrong.

a = [(a+b)^n + (a-b)^n] / 2
b = [(a+b)^n - (a-b)^n] / 2

-b = - [(a+b)^n - (a-b)^n] / 2

I'll take a1,1 as example.

A11,1 = a A21,1 = a2-b2
A31,1= a(a2-3b2)

However when I use the general expression a = [(a+b)^n + (a-b)^n] / 2
I'm getting:

A1 = a A2 = a2+b2
A3= a(a2+3b2)

I've tried a few variations with trying to incorporate a negative but with no success.
Any help greatly appreciated

5. Aug 18, 2009

### jacobrhcp

That guys answer is very illuminating and useful, you should read up on it as soon you know what egenvectors are.

the constants in the matrix are just coefficients appearing also in (a+b)^n, where the even powers of b appear on the top-left sides and the uneven powers on the top-right side.

I'll write powers of A as a vector of the top two coefficients for my convinience:

(a+b)^n can be expanded using newtons binomium

A=(a,-b) (from a+b, left 1st, right 2nd)
A^2=(aa - bb, -2ab) (these are coefficients from (a+b)^2, left the 1st and 3d, right the 2nd)
etc...
writing n over k as n_k, again for convenience, we get:

A^n= (a^n + (-1)*(n_n-2)*(a^n-2)*b^2 + ... + (-1^k-1)*(n_n-2k)*(a^n-2k)*b^2k, n*(a^n-1)*b + ... + (-1)^k*(a^n-2k-1)*(n_n-2k-1)*b^2k-1)

which can be easier notated with a sum symbol, but I don't like LaTeX

You can do induction to proof this is correct, I'm convinced it is.

=)

Last edited: Aug 19, 2009