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Prove that the left and right shifts on $l_{2}$ have no polar decomposition (i.e. $UP$ where $U$ is unitary and $P$ is positive).

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Left Shift

Denote for shorthand: $$1_0(0,x_1,\ldots):=(0,x_1,\ldots)$$

For the modulus: $$L^*L=RL=1_0\implies|L|=1_0$$

For the argument: $$L=U|L|\implies U=1_0,1,\ldots$$

So it admits one.

Right Shift

For the modulus: $$R^*R=LR=1\implies|R|=1$$

For the argument: $$R=U|R|\implies U=R$$

So it admits none.

Reference

For much more details: Polar Decomposition

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  • In your first line do you mean $1_0(\color{red}{x_1},x_2,\ldots):=(0,\color{red}{x_2},\ldots)$? – ViktorStein Mar 07 '20 at 12:02
  • I thought we had $$R L (x_1, x_2, x_3, \ldots) = R(x_2, x_3, \ldots) = (0, x_2, x_3, \ldots),$$ implying that $$|L| (x_1, x_2, x_3, \ldots) = RL (x_1, x_2, x_3, \ldots) = (0, x_2, x_3, \ldots)$$ Then we have $$L (x_1, x_2, x_3, \ldots) = (x_2, x_3, \ldots) \overset{!}{=} (U | L |) x = U (0, x_2, x_3, \ldots),$$ implying $U(x_1, x_2, \ldots) = (x_2, x_3,\ldots)$ What you mean by "$U = 1_0, 1, \ldots$"? – ViktorStein Mar 07 '20 at 12:13
  • The sequences in this post were meant to start with $x_0$, i.e. $(x_0,x_1,x_2,\ldots)$. – C-star-W-star Mar 08 '20 at 21:26