# What is the difference between isometric and unitary operators 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. Aug 16 '14 at 14:43
• I think unitary operator guarantees invertibility, i.e. U* = inv(U); however, isometric operator not. Aug 16 '14 at 17:26
• @sleeve: That's right. You succinctly answered your own question a few minutes before I posted. 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.

I'm not sure what you mean by "isomorphic". One notion of equivalence of linear transformations is similarity; but a surjective operator is never similar to a nonsurjective operator. A stronger notion is unitary equivalence, i.e., similarity induced by a unitary transformation (since these are the isometric isomorphisms of Hilbert space), which again cannot happen between a nonunitary isometry and a unitary operator (or between any nonunitary operator and a unitary operator).

• I cited your answer in my question here math.stackexchange.com/q/2091816. I'd appreciate your feedback. Jan 10 '17 at 13:39
• 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). Jul 7 '18 at 6:09
• @tparker I don't think you need to work in the eigenbasis. Use $x=e_i$ to get $(O^TO)_{ii}=1$, and then $x=e_i+e_j$ to see that $(O^T O)_{ij}+(O^T O)_{ji}=2(O^T O)_{ij}=0$
– glS
Apr 26 '20 at 8:40
• If $||Tx|| = ||x|| \ \ \ \forall x$, does that imply $||T||=1 \ \ \ \forall$ isometric operators? Jan 19 at 7:10
• @ric.san: Yes isometric operators do have operator norm $1$. Mar 13 at 5:02

In finite dimensions, there is a straightforward characterisation of isometries and unitaries in terms of their matrix representations.

The basic observation is that $$U^*U=I$$ means that the columns of (the matrix representation of) $$U$$ are orthonormal, while $$UU^*=I$$ means that the rows of $$U$$ are orthonormal.

A unitary $$U$$ is a matrix whose columns (equivalently, rows) form an orthonormal basis. This is equivalent to both its columns and rows being orthonormal. An isometry, on the other hand, only requires that the columns are orthonormal, but not that they form a basis.

In summary, the distinction between the two objects can be stated as follows: an isometry is a matrix whose columns are orthonormal, while a unitary is squared matrix whose columns are orthonormal.