Thread: Abstract Algebra Matrix Formulas

Find a formula for the n-th power of any 2 x 2 upper-triangular real matrix,
a...b
0...d
where n is a positive integer, and verify the formula by induction on n. Then, assuming a and d are nonzero, show the matrix is invertible and determine if there is a single formula that works for all integers n (i.e., when n < 0 too).

So, first in order to derive a general formula for the matrix, I did out the multiplication. The second term in the first row is causing me problems. I find that the first term in the first row is a^n, and the last term in the last row is d^n, but the second term in the first row is causing problems. Below are my results for the second term for each n:
n=1 : b
n=2: ab + bd
n=3: a^2b + abd + bd^2
n=4: a^3b + a^2bd + abd^2 + bd^3

I see a pattern, but am not sure how to represent the pattern in terms of n, so any help with this would be appreciated. I was thinking of something along the lines of a^(n-1)b + a^(n-2)bd... but I'm not really sure that's the way to go. Also, any other tips/advice while working on the rest of the problem would be awesome, and if there is a formula that works for negative integers of n as well.

Take the b out and things look a little clearer.






The expression in brackets can be written neatly as sum.

Also, .

Originally Posted by emakarov

Also, .

I don't fully understand this. My experience with sum notation is brief. Could you explain this a little? When I look at that sum, I see it as a^n-0d⁰+a^n-1d¹...+ a^n-ndⁿ. Is that incorrect? Is that the correct way of representing the summation that's happening as n increases? I don't understand where you are getting (aⁿ⁺¹-dⁿ⁺¹)/(a-d). Maybe I'm missing something?

Example

Originally Posted by TheHowlingLung

When I look at that sum, I see it as a^n-0d⁰+a^n-1d¹...+ a^n-ndⁿ.

This is correct.

Originally Posted by TheHowlingLung

I don't understand where you are getting (aⁿ⁺¹-dⁿ⁺¹)/(a-d).

The easiest way to prove this is to multiply (a - d) by (aⁿd⁰+a^n-1d¹...+ a⁰dⁿ); most terms cancel out. See also here and here.

Thanks guys! I appreciate the help.

Originally Posted by emakarov

This is correct.

The easiest way to prove this is to multiply (a - d) by (aⁿd⁰+a^n-1d¹...+ a⁰dⁿ); most terms cancel out. See also here and here.

Edit: The first link does not work, but it's in the Math Help Boards forum. The part of the URL after the site name is correct.