Is there any quick argument to show that every non-nilpotent matrix $A\in \mathbb C^{2 \times2}$ has a square root? Just the existence without computing it.
Knowing that $A\in \mathbb C^{2 \times2}$ is non-nilpotent basically tells us that at least one eigenvalue is non-zero. But it's far from being diagonalizable.
My text haven't introduced factorisations based on orthogonality yet. So I'm expecting the proof to be based on Jordan form.
The case where $A$ is diagonalizable is probably know to you.
Assume $A$ isn't diagonalizable. Then, since it is a $2\times 2$ matrix, it's jordan normal form is $\color{grey}{J=}\begin{bmatrix} \lambda & 1\\ 0 & \lambda\end{bmatrix}$, where $\lambda$ is $A$'s only eigenvalue.
It suffices to find a square root of $J$. Assume it exists and denoted it by $\sqrt J$.
Take the ansatz $\sqrt J=\begin{bmatrix} \sqrt \lambda & a\\ b & \sqrt \lambda \end{bmatrix}$, where $\sqrt \lambda$ denotes a square root of $\lambda$.
Then you want $\sqrt J^2=J$, that is $\begin{bmatrix} \lambda +ab & 2a\sqrt \lambda \\ 2b\sqrt \lambda & ab+\lambda \end{bmatrix}=\begin{bmatrix} \lambda & 1\\ 0 & \lambda\end{bmatrix}$, yielding $a=\dfrac 1{2\sqrt \lambda}\land b=0$.
This gives you four different possibilities for $\sqrt J$.
Edit: In retrospective taking $b\neq 0$ in the ansatz is a bit silly. One might as well try to look for upper triangular square roots from the start.
This answer is relevant.
The result is trivial when $A$ is diagonalisable. So, we only need to consider the case where $A$ is non-diagonalisable and non-nilpotent. Hence $A$ has two equal but nonzero eigenvalues $\lambda$.
By Cayley-Hamilton theorem, $A^2=\operatorname{tr}(A)-\det(A)I=2\lambda A-\lambda^2I$. Therefore $A=\frac1{4\lambda}(A+\lambda I)^2$.
Alternatively, as $A$ is invertible, it has a matrix logarithm. Therefore $A=(e^{(\log A)/2})^2$.
