# Difference between an isometric operator and a unitary operator on a Hilbert space

It seems that both isometric and unitary operators on a Hilbert space have the following property:

$$U^*U = I$$ ($$U$$ is an operator and $$I$$ is an identity operator, $$^*$$ is a binary operation.)

What is the difference between isometry and unitary? Which one is more general, or are they the same? Are they isomorphic?

• Unitary is complex isometry. – Shuchang Aug 16 '14 at 14:43
• I think unitary operator guarantees invertibility, i.e. U* = inv(U); however, isometric operator not. – sleeve chen Aug 16 '14 at 17:26
• @sleeve: That's right. You succinctly answered your own question a few minutes before I posted. – Jonas Meyer Aug 16 '14 at 17:35

An isometric operator on a (complex) Hilbert space is a linear operator that preserves distances. That is, $T$ is an isometry if (by definition) $\|Tx-Ty\|=\|x-y\|$ for all $x$ and $y$ in the space. By linearity, this is equivalent to $\|Tx\|=\|x\|$ for all $x$. Because of the definition of the norm in terms of the inner product and the definition of adjoint operators, this is equivalent to $\langle T^*Tx,x\rangle=\langle x,x\rangle$ for all $x$. This implies that $T^*T=I$. Conversely, if $T^*T=I$, you can show that $T$ is an isometry (this direction is easier).
A unitary operator $U$ does indeed satisfy $U^*U=I$, and therefore in particular is an isometry. However, unitary operators must also be surjective (by definition), and are therefore isometric and invertible. They are the isometric isomorphisms on Hilbert space. One way to characterize them algebraically is to say that $U$ is a unitary if $U^*U=UU^*=I$.
On infinite dimensional Hilbert spaces (unlike in finite dimensional cases), there are always nonunitary isometries. For example, on $\ell^2$, the operator sending $(a_0,a_1,a_2,a_3,\ldots)$ to $(0,a_0,a_1,a_2,\ldots)$ is a nonunitary isometry.
• Interestingly, in the real case the argument in your first paragraph needs to be (slightly) more complicated: going from $⟨O^T O x, x⟩=⟨x, x⟩$ to $O^T O = I$ requires a stronger argument than in the complex case (e.g., using the fact that $O^T O$ is symmetric and therefore diagonalizable, and going into an eigenbasis). – tparker Jul 7 '18 at 6:09