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?

  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Aug 14 at 17:07

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