Square root of 2x2 matrix using basic algebra

We want to find some matrix \(B\) where $ B = $ or, equivalently, $ A = B^2 $. We take

\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

\[ B = \begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix} \]

Expand \(B\) and equate terms.

from sympy import *
init_printing()
A,a,b,c,d,e,f,g,h = symbols('A a b c d a\' b\' c\' d\'')

A = Matrix([[a,b], [c,d]])
B = Matrix([[e,f], [g,h]])

e1 = Eq(A,B**2)
e1

\[\left[\begin{matrix}a & b\\c & d\end{matrix}\right] = \left[\begin{matrix}a'^{2} + b' c' & a' b' + b' d'\\a' c' + c' d' & b' c' + d'^{2}\end{matrix}\right]\]

solve(e1, (a,b,c,d))

\[\left \{ a : a'^{2} + b' c', \quad b : b' \left(a' + d'\right), \quad c : c' \left(a' + d'\right), \quad d : b' c' + d'^{2}\right \}\]

We can calculate the determinant \(A\) and \(B^2\) (confirming that \(det(B^2) = det(B)^2\))

Eq(A.det(), factor((B**2).det()))

\[a d - b c = \left(a' d' - b' c'\right)^{2}\]

We can also calculate the trace of \(A\) and \(B^2\) \[ trace(A) = trace(B^2)\]

Eq(A.trace(), (B**2).trace())

\[a + d = a'^{2} + 2 b' c' + d'^{2}\]

But from the equation for the determinant we know that \[ ad-bc=(a'd'-b'c')^2 \] or \[b'c' = a'd'-\sqrt{ad-bc}\] so \[ a+d = a'^2 + 2a'd' \pm 2\sqrt{ad-bc} +d'^2\] \[a+d = (a'+d')^2 \pm 2\sqrt{ad-bc}\] so \[ trace(B) = a'+d' = \pm \sqrt{a+d \pm 2\sqrt{ad-bc}} = \pm \sqrt{trace(A) \pm 2\sqrt{det(A)}}\]

Giving 4 solutions. Using these two relations we can calculate the expressions for \(a'\),\(b'\), \(c'\) and \(d'\) in terms of \(a\), \(b\), \(c\) and \(d\). For example, \[b=b'(a'+d')\] \[b=b'(\pm \sqrt{a+d \pm 2\sqrt{ad-bc}})\] \[b'=\pm \frac{b}{\sqrt{a+d \pm 2\sqrt{ad-bc}}}\] etc…