Two projections $P,Q$ are unitarily equivalent if and only if $$dim(randP)=dim(ranQ)$$ $$dim(kerP)=dim(kerQ)$$

How can we show this? One directionn seems easy:

If $P$ and $Q$ are unitarily eqv, then $PU=UQ$ where $U$ is an isomorphism,hence preserves the dimension of the Hilbert space. Therefore, $dimranP=dimranQ$.

How to show the other direction? Please help me with this.

  • $\begingroup$ It's false.${}$ $\endgroup$ – Olivier Bégassat May 7 '15 at 13:20
  • $\begingroup$ Are you assuming your projections are orthogonal projections (equivalently, that they are self-adjoint)? $\endgroup$ – Daniel Fischer May 7 '15 at 13:23
  • $\begingroup$ @DanielFischer Yes, aren't projections always self adjoint? $\endgroup$ – L.G May 7 '15 at 13:38
  • $\begingroup$ @OlivierBégassat Maybe you could elaborate a little on your answer? $\endgroup$ – L.G May 7 '15 at 13:38
  • 1
    $\begingroup$ A projection is just a linear operator $P$ such that $P^2 = P$, it need not even be continuous, let alone self-adjoint. But it's not quite uncommon that when considering projections on Hilbert spaces, one restricts the attention to self-adjoint projections, and omits the qualification, since one doesn't consider other projections. That's why I asked. (And it's even more common to omit the "continuous" qualification when one only considers continuous projections.) $\endgroup$ – Daniel Fischer May 7 '15 at 13:47

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