Square Root of a Matrix

[scikidus]

I've been playing with idea for a while, but only on 2*2 maticies. Have fun.

Given a square n-by-n matrix B, find a square n-by-n matrix A such that A*A = B.

I've only toyed with n=2, where I proved that |B| = (|A|)^2:

Spoiler:

Let A = \left( \begin{array}{cc}
a & b\\
c & d\\ \end{array} \right) and B = \left( \begin{array}{cc}
w & x\\
z & y\\ \end{array} \right).

Since A*A = B, we know that

w = a^2 + bc
x = b(a+d)
y = c(a+d)
z = d^2 + bc

|B| = wz-xy = (a^2+bc)(d^2+bc) - bc(a+d)^2 =
(ad)^2 + bca^2 + bcd^2 + (bc)^2 - bca^2 - 2(ad)(bc) - bcd^2 =
(ad)^2 - 2(ad)(bc) + (bc)^2 =
(ad - bc)^2
= (|A|)^2

QED.

Now I've got a few questions.

1. Does this rule of |B| = (|A|)^2 hold up for larger square matricies with n>2?
2. Is there a general equation for finding A just from B? A general equation for finding how many possible A's there are?
3. What happens when you try cube roots? n-th roots?

If you want a (very) easy problem to try, find A for
B = \left( \begin{array}{cc}
18 & 14\\
7 & 11\\ \end{array} \right)

[antonfire]

1. Does this rule of |B| = (|A|)^2 hold up for larger square matricies with n>2?

Of course. Part of the "point" of determinants is that det(A*B) = det(A)*det(B).

2. Is there a general equation for finding A just from B? A general equation for finding how many possible A's there are?

It's a lot easier if you find the Jordan normal form of the matrix. Then you can just take "roots" of the blocks. Unfortunately to do this you have to work with the complex numbers, and then you have to see if the roots you get give you real matrices. But you sort of do that when you're taking roots of numbers anyway.

Edit: actually, it is a bit more complicated than it might look at first since the bits corresponding to the same eigenvalue can "interact" to give you more roots than you'd expect. For example -I has square roots with real coefficients.

Another way is to note that you can just use the power series for log and exp (and even sqrt) with matrices and things sometimes work out fine.

3. What happens when you try cube roots? n-th roots?

More or less the same thing.

[Nitrodon]

B = \left( \begin{array}{cc}
-\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}}\\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\ \end{array} \right)
\left( \begin{array}{cc}
4 & 0\\
0 & 25\\ \end{array} \right)
\left( \begin{array}{cc}
-\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}}\\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\ \end{array} \right), where the first and last matrices in that product are inverses. (Coincidentally, they are also both the same matrix in this case.)

Hence A = \left( \begin{array}{cc}
-\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}}\\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\ \end{array} \right)
\left( \begin{array}{cc}
\pm2 & 0\\
0 & \pm5\\ \end{array} \right)
\left( \begin{array}{cc}
-\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}}\\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\ \end{array} \right) =
\pm\left( \begin{array}{cc}
4 & 2\\
1 & 3\\ \end{array} \right) or \pm\frac{1}{3}\left( \begin{array}{cc}
8 & 14\\
7 & 1\\ \end{array} \right)

If you have a repeated eigenvalue, the general solution becomes more complicated.

[t0rajir0u]

Deciding when a matrix over the complex numbers has a square root is actually fairly delicate. To understand what's going on you should first learn about Jordan normal form and see if you can follow the discussion here and here.

A result that is easy to state is that any diagonalizable matrix has a square root: just take square roots of its eigenvalues.