# Alternative proofs invertible matrix/operator $T$ has square root over complex field using polar decomposition

My textbook gives the following proof: (i) $$I+N$$ where $$N$$ is nilpotent operator has square root (ii) $$T$$ may be decomposed into $$T|_{G(\lambda_i)}=\lambda_i I+N_i$$ (restriction to generalized eigenspace).

Another proof I think might work is to use polar decomposition $$T=S \sqrt{T^*T}$$ where the isometry $$S$$ can be diagonalized as it is a normal operator wrt. complex field and thus has a square root. And $$\sqrt{T^*T}$$ is positive operator which means it has a unique positive square root. The problem is now that invertibility has not been used and I cannot show the two square roots are commutative. Any suggestion how to proceed from here?

• In general, $S$ and $\sqrt{T^*T}$ will commute if and only if the original matrix $T$ commutes with its adjoint $T^*$. I think that your method of proof fails. – Omnomnomnom Aug 14 at 17:07