# Infinite sum of matrices

1. Jan 29, 2009

### Hannisch

1. The problem statement, all variables and given/known data
http://img144.imageshack.us/my.php?image=matrixkx2.jpg

Find the sum of the infinte series: I3+A+A2+A3...

2. Relevant equations
A*I=I*A=A
A*A-1=A-1*A=I

And probably some other things as well.

3. The attempt at a solution
This is my (our) solution to this problem, could someone check and see if we actually can do all of this mathematically? Matrices are .. fairly new. (Say, three weeks or so.)

S=the infinite sum

S=I3+A+A2+A3...

AS=IA+A2+A3...

S - AS=I3+A+A2+A3... - IA+A2+A3...

S - AS=I

S(I - A)=I

S(I - A)(I - A)-1=I(I - A)-1

SI=(I - A)-1

S=(I - A)-1

From this we just calculate I - A, which is simple enough, and then take (I-A)*X=I, X being our inverse matrix with entries something like a through i. Just doing matrix multiplication we get nine expressions for the different nine terms and in the end get a matrix, which is then the inverse matrix of (I - A).

Is this a good solution for this problem?

2. Jan 29, 2009

### Staff: Mentor

Looks good to me.

3. Jan 29, 2009

### CompuChip

The final answer is ok, so probably so is the rest. I do have some remarks though...

You need to be careful with subtracting infinite sums (such as S and AS) from each other, although in this case it's fairly obvious that it works out.

More importantly, I noticed some sloppiness in the ordering of the matrices. For example, in the first step, you write A S = I A + ....
Actually you are multiplying S from the left by A, so if S starts with I + ... then you would get A I + .... Of course, since A I = I A this doesn't matter in this case, but you might as well do it correctly from the start. The same happens later on, when you write
S - A S = S(I - A).
Actually, when you work out the brackets, this says S - A S = S - S A, which would mean that A S = S A. Of course in this case, because S only consists of products of A's and the identity matrix, this is true; but in general it is not. What you should have written is of course S - A S = (I - A) S.

Finally, although I think your solution is rather neat, I would like to give you another approach to this problem. After multiplying out some of the matrices by hand, you will start to notice a pattern, namely that every even power looks like
$$A^{2n} = \begin{pmatrix} 1/2^{2n} & 0 & 0 \\ 0 & 1/2^{2n} & 0 \\ 0 & 0 & 1/2^{2n} \end{pmatrix}$$
and the odd powers look like
$$A^{2n+1} = \begin{pmatrix} 0 & 1/2^{2n+1} & 0 \\ 1/2^{2n + 1} & 0 & 0 \\ 0 & 0 & 1/2^{2n} \end{pmatrix}$$

Since summing matrices means summing the entries, the (3, 3) entry for example then simply becomes a geometric series
$$\sum_{n = 0}^\infty \frac{1}{2^n} = 2$$
and similarly you can write down a series for the other four non-zero entries.
Just wanted to show you the method for future reference

4. Jan 29, 2009

### Dick

That's a good solution. The answer is the inverse of (I-A).

5. Jan 29, 2009

### CompuChip

By the way, notice this funny statement: you probably know the geometric series
$$\sum_{x = 0}^\infty x^n = \frac{1}{1 - x} = (1 - x)^{-1}$$
(actually, this is true for |x| < 1).

If you write A instead of x, you get
$$\sum_{x = 0}^\infty A^n = \frac{1}{1 - A} = (1 - A)^{-1}$$
where of course the "fraction" notation doesn't make sense for a matrix, but if you write it in the second form and read 1 as the identity, then this is exactly like the geometric series.

Important: don't want you to think that you could have used this to derive the answer, the geometric series is technically only valid for real numbers -1 < x < 1; I just wanted to point out a striking similarity to show that sometimes matrices can be just like numbers (however, sometimes they are not, so I strongly urge you never to take such shortcuts like the one above.)

6. Jan 29, 2009