How to calculate #A^n#,with #A=((1,2,3),(0,1,2),(0,0,1))#?
2 Answers
Explanation:
Let us look at the first few powers of
#A^0 = ((1, 0, 0),(0, 1, 0),(0, 0, 1))#
#A^1 = ((1, 2, 3),(0, 1, 2),(0, 0, 1))#
#A^2 = ((1, 2, 3),(0, 1, 2),(0, 0, 1))((1, 2, 3),(0, 1, 2),(0, 0, 1)) = ((1, 4, 10),(0, 1, 4),(0, 0, 1))#
#A^3 = ((1, 2, 3),(0, 1, 2),(0, 0, 1))((1, 4, 10),(0, 1, 4),(0, 0, 1)) = ((1, 6, 21),(0, 1, 6),(0, 0, 1))#
#A^4 = ((1, 2, 3),(0, 1, 2),(0, 0, 1))((1, 6, 21),(0, 1, 6),(0, 0, 1)) = ((1, 8, 36),(0, 1, 8),(0, 0, 1))#
It looks like the form of
#A^n = ((1, 2n, a_n),(0, 1, 2n),(0, 0, 1))#
with the
#color(blue)(3), 10, 21, 36,..." "# for#n=1,2,3,4,...#
Writing down the sequence of differences between consecutive terms, we get:
#color(blue)(7), 11, 15#
Writing down the sequence of differences between those differences, we get:
#color(blue)(4), 4#
Having reached a constant sequence, we can use the first term of each sequence to give a direct formula for
#a_n = color(blue)(3)/(0!)+color(blue)(7)/(1!)(n-1)+color(blue)(4)/(2!)(n-1)(n-2)#
#color(white)(a_n) = 3+7n-7+2n^2-6n+4#
#color(white)(a_n) = 2n^2+n#
So it looks like:
#A^n = ((1, 2n, 2n^2+n),(0, 1, 2n),(0, 0, 1))#
We can prove this by induction:
Case
#((1, 2(color(blue)(0)), 2(color(blue)(0))^2+(color(blue)(0))),(0, 1, 2(color(blue)(0))),(0, 0, 1)) = ((1, 0, 0),(0, 1, 0),(0, 0, 1)) = A^0#
Induction step
Suppose:
#A^n = ((1, 2n, 2n^2+n),(0, 1, 2n),(0, 0, 1))#
Then:
#((1, 2(color(blue)(n+1)), 2(color(blue)(n+1))^2+(color(blue)(n+1))),(0, 1, 2(color(blue)(n+1))),(0, 0, 1)) = ((1, 2n+2, 2n^2+5n+3),(0, 1, 2n+2),(0, 0, 1))#
#color(white)(((1, 2(n+1), 2(n+1)^2+(n+1)),(0, 1, 2(n+1)),(0, 0, 1))) = ((1, 2, 3),(0, 1, 2),(0, 0, 1))((1, 2n, 2n^2+n),(0, 1, 2n),(0, 0, 1)) = A A^n = A^(color(blue)(n+1))#
Remarks
Given that we now have a formula for
Using our formula:
#A^(1/2) = ((1, 1, 1),(0,1,1),(0,0,1))#
We find:
#((1, 1, 1),(0, 1, 1),(0, 0, 1))((1, 1, 1),(0, 1, 1),(0, 0, 1)) = ((1, 2, 3), (0, 1, 2), (0, 0, 1))#
as expected.
This would be quite interesting to generalise to other forms of matrix. Does it give a generic way of finding a square root of a matrix?
See below.
Explanation:
Matrix
and
There is a difference equation associated to this relationship
The difference equation solution is
Here