# Hilbert Space Isomorphism is Equivalence Relation

## Theorem

Hilbert space isomorphism is an equivalence relation.

## Proof

Checking the three defining properties of an equivalence relation in turn:

### Reflexivity

For any Hilbert space $H$, we have the identity map $\operatorname{id}: H \to H$.

Identity Mapping is Linear yields $\operatorname{id}$ to be a linear map.

Also, Identity Mapping is Bijection ensures that $\operatorname{id}$ is a surjection.

Lastly, for any $g, h \in H$, we have:

- $\left\langle{g, h}\right\rangle = \left\langle{\operatorname{id} \left({g}\right), \operatorname{id} \left({h}\right)}\right\rangle$

Therefore, the three defining properties for a Hilbert space isomorphism are satisfied by $\operatorname{id}$.

Hence, any Hilbert space $H$ is isomorphic to itself.

It follows that Hilbert space isomorphism is reflexive.

$\Box$

### Symmetry

Let $H, K$ be isomorphic Hilbert spaces, and let $U: H \to K$ be a Hilbert space isomorphism.

From Hilbert Space Isomorphism is Bijection and the definition of bijection, it follows that $U$ has a two-sided inverse, $U^{-1}: K \to H$.

Inverse of Linear Map is Linear yields $U^{-1}$ to be a linear map as well.

Bijection iff Inverse is Bijection yields that $U^{-1}$ is also a surjection.

For $k, l \in K$, the computation $\left\langle{k, l}\right\rangle_K = \left\langle{U U^{-1}k, U U^{-1}l}\right\rangle_K = \left\langle{U^{-1}k, U^{-1}l}\right\rangle_H$ shows that $U^{-1}$ preserves the inner product (where in the last step, the fact that $U$ is an isomorphism is used).

Hence, $U^{-1}$ is an isomorphism, so $K$ and $H$ are also isomorphic.

It follows that Hilbert space isomorphism is symmetric.

$\Box$

### Transitivity

Let $H, K, L$ be Hilbert spaces, and let $U: H \to K, V: K \to L$ be Hilbert space isomorphisms.

Consider the composition $VU : H \to L$.

From Composition of Linear Maps is Linear, $VU$ is a linear map.

Composite of Surjections is Surjection yields $VU$ to be a surjection.

Lastly, for $g, h \in H$, we have:

- $\left\langle{g, h}\right\rangle_H = \left\langle{Ug, Uh}\right\rangle_K = \left\langle{VUg, VUh}\right\rangle_L$

Hence $VU$ is a Hilbert space isomorphism, and $H$ and $L$ are isomorphic.

It follows that Hilbert space isomorphism is transitive.

$\Box$

Having verified these three conditions, it follows that Hilbert space isomorphism is an equivalence relation,

$\blacksquare$

## Sources

- 1990: John B. Conway:
*A Course in Functional Analysis*: $\S I.5$