Question

How to find `X` knowing that `X^2=((4,1),(0,4))`?

Answer 1

`X = +-((2, 1/4), (0, 2))`

Explanation:

This question is asking something I have wondered about myself: How do you find a square root of a matrix?

Let's try it the naive way, just squaring a matrix then solving the resulting simultaneous equations:

Suppose:

`((4, 1), (0, 4)) = ((a, b), (c, d))((a, b), (c, d))`

`color(white)(((4, 1), (0, 4))) = ((a^2+bc, b(a+d)), (c(a+d), bc+d^2))`

Note that if we put `c=0` then `bc = 0`, so:

`a=+-sqrt(4) = +-2`

`d=+-sqrt(4) = +-2`

Let's choose `a=d=2`

Then:

`b(a+d) = 1` and hence `b=1/4`

So a solution is:

`X = ((2, 1/4), (0, 2))`

Note that `(-X)^2 = X^2`, so `-((2, 1/4), (0, 2))` is also a solution.

Note that we cannot give `a` and `d` opposite signs, since that would make `a+d = 0` and we would not be able to make `b(a+d) = 1`.

Answer 2

`X = +-((2, 1/4), (0, 2))`

Explanation:

Here's another method using an interesting property of matrices.

Note that:

`((1, a), (0, 1))((1, b), (0, 1)) = ((1, a+b),(0, 1))`

Hence in particular we find:

`((1, a), (0, 1))^n = ((1, na), (0, 1))`

`((1, a), (0, 1))^2 = ((1, 2a), (0, 1))`

and we can define a "principal" square root:

`((1, a), (0, 1))^(1/2) = ((1, a/2), (0, 1))`

and non-principal square root:

`-((1, a), (0, 1))^(1/2) = -((1, a/2), (0, 1))`

So in our example:

`((4, 1), (0, 4))^(1/2) = (4 ((1, 1/4), (0, 1)))^(1/2) = 4^(1/2) ((1, 1/4), (0, 1))^(1/2) = 2 ((1, 1/8), (0, 1)) = ((2, 1/4), (0, 2))`

So our solutions for `X^2 = ((4, 1), (0, 4))` are:

`X = +-((2, 1/4), (0, 2))`