Question

How to calculate `A^n`,with `A=((1,2,3),(0,1,2),(0,0,1))`?

Answer 1

`A^n = ((1, 2n, 2n^2+n),(0, 1, 2n),(0, 0, 1))`

Explanation:

Let us look at the first few powers of `A` to find the pattern:

`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` is something like:

`A^n = ((1, 2n, a_n),(0, 1, 2n),(0, 0, 1))`

with the `a_n`'s forming the sequence:

`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`:

`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 `n=0`

`((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))`

`color(white)()`
Remarks

Given that we now have a formula for `A^n`, what about non-integer values of `n`?

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?

Answer 2

See below.

Explanation:

Matrix `A` characteristic polynomial is

`p(lambda)=1 - 3 lambda + 3 lambda^2 - lambda^3 =- (lambda-1)^3`

and `A` obeys this characteristic equation so

`A^3=3A^2-3A+I`

There is a difference equation associated to this relationship

`a_(n+3)=3a_(n+2)-3a_(n+1)+a_n`

The difference equation solution is

`a_n=C_0+nC_1+n^2C_2` so finally

`A^n = C_0+nC_1+n^2C_2`

Here

`C_0 = ((1,0,0),(0,1,0),(0,0,1))`
`C_1=((0, 2, 1),(0, 0, 2),(0, 0, 0))`
`C_2=((0,0,2),(0,0,0),(0,0,0))`