Existence of square root matrix $B \in \mathbb{C}^{2\times 2}$ for any $A \in \mathbb{C}^{2\times 2}$, where $A^2\neq 0$

Question

I am trying to prove that for any $A \in \mathbb{C}^{2\times 2}$ with $A^2\neq 0$, there exists $B \in \mathbb{C}^{2\times 2}$ with $BB=A$.

I have tried the approach of a general matrix A andB with variable entries $$B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ $$A = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}$$

and assuming $BB=A$, I get the equations

$$a^2+bc=\alpha$$ $$b(a+d)=\beta$$ $$c(a+d)=\gamma$$ $$d²+cb=\delta$$

However, here I am stuck since I do not know whether any of those variables is $0$, so I cannot operate with those equations.

I have seen a solution on Wikipedia, however to me it seems to fall from the sky, especially the restrictions it makes.

I have also found a thread, which shows that without the restriction $A^2\neq0$ this statement is false, however I fail to see how this is the critical restriction.

Explanations, clarifications or hints on any of the things I mentioned are most welcome.

Answer 1

Either $A$ is similar to a diagonal matrix, of which we can easily find a square root. Or it is similar to an upper triangular matrix $\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$. Note that $x^2=\lambda$ implies $\begin{pmatrix}x&y\\0&x\end{pmatrix}^2=\begin{pmatrix}\lambda&2xy\\0&\lambda\end{pmatrix}$, so $y=\frac1{2x}$ gives us a solution (as $x\ne0$).

Answer 2

We have the equations \begin{eqnarray*} a^2+bc= A \\ b(a+d)=B \\ c(a+d)=C \\ bc+d^2=D \end{eqnarray*} Multiply the first equation by $(a+d)^2$ and use the second & third we have \begin{eqnarray*} a^2(a+d)^2 +BC=A(a+d)^2 \\ d= -a +\sqrt{\frac{BC}{A-a^2}}. \end{eqnarray*} Now subtract the first & the fourth \begin{eqnarray*} a^2-d^2=A-D \\ \end{eqnarray*} Substitute for $d$ and we have \begin{eqnarray*} a^2-A+D= \left( -a +\sqrt{\frac{BC}{A-a^2}} \right)^2 \\ D-A-\frac{BC}{A-a^2}=-2a \sqrt{\frac{BC}{A-a^2}} \end{eqnarray*} Square this & we have a quadratic in $a^2$ \begin{eqnarray*} a^4((D-A)^2+4BC)+a^2(-2A(D-A)^2-2BC(A-D)-4ABC)+A^2(D-A)^2-2ABC(D-A)+B^2C^2=0 \end{eqnarray*} Note that this has discriminant $ \Delta=4B^2C^2(AD-BC)$. This gives \begin{eqnarray*} a^2= \frac{A(A-D)^2+BC(3A-D) \mp 2BC \sqrt{AD-BC}}{(a-d)^2+4BC} \\ =A-\frac{BC}{A+D \pm 2 \sqrt{AD-BC}} \end{eqnarray*} Once the dust has settled ... \begin{eqnarray*} \sqrt{\left[ \begin{array}{cc} A & B \\ C & D\\ \end{array} \right]}=\left[ \begin{array}{cc} \sqrt{A-\frac{BC}{A+D \pm 2 \sqrt{AD-BC}} } & \frac{B}{\sqrt{A+D \pm 2 \sqrt{AD-BC}}} \\ \frac{C}{\sqrt{A+D \pm 2 \sqrt{AD-BC}}} & \sqrt{D-\frac{BC}{A+D \pm 2 \sqrt{AD-BC}} } \\ \end{array} \right] \end{eqnarray*}