A non-nilpotent matrix $A\in \mathbb C^{2 \times2}$ has a square root 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 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 $A=(e^{(\log A)/2})^2$.
