Suppose $\( H \)$ is a Hilbert space, $\( {T}_{1} \)$ and $\( {T}_{2} \)$ are closed orthogonal subspaces, and $\( T \)$ is a closed subspace such that $\( {T}_{1}\subseteq T \)$. Then $\( {T}_{1}\oplus {T}_{2} \)$ is closed, and there exists exactly one closed subspace, denoted $\( T\ominus {T}_{1} \)$, such that $\( \mathopen{}\left(T\ominus {T}_{1}\right)\mathclose{}\subseteq T \)$, $\( \mathopen{}\left(T\ominus {T}_{1}\right)\mathclose{}\perp {T}_{1} \)$, and $\( T= {T}_{1}\oplus \mathopen{}\left(T\ominus {T}_{1}\right)\mathclose{} \)$. This subspace is called the orthogonal complement of $\( {T}_{1} \)$ with respect to $\( T \)$.

For $\( H \)$ a Hilbert space and $\( T\subseteq H \)$ closed, $\( H\ominus T= {T}^{\perp} \)$.

The external direct sum of a finite number of Hilbert spaces $\( {H}_{1} \)$, …, $\( {H}_{n} \)$ with inner products $\( \mathopen{}\left\langle{}\cdot, \cdot\right\rangle\mathclose{}_{1} \)$, …, $\( \mathopen{}\left\langle{}\cdot, \cdot\right\rangle\mathclose{}_{n} \)$, written $\( {H}_{1}\oplus \dotsb\oplus {H}_{n} \)$, is the set of $\( n \)$-tuples $\( \mathopen{}\left\{\, \mathopen{}\left({x}_{1}, \dotsc, {x}_{n}\right)\mathclose{}\,\middle\vert\, , {x}_{i}\in {H}_{i}, \,\right\}\mathclose{} \)$.

The direct sum of Hilbert spaces is a Hilbert space with

1. $\( \mathopen{}\left({x}_{1}, \dotsc, {x}_{n}\right)\mathclose{}+\mathopen{}\left({y}_{1}, \dotsc, {y}_{n}\right)\mathclose{}= \mathopen{}\left({x}_{1}+{y}_{1}, \dotsc, {x}_{n}+{y}_{n}\right)\mathclose{} \)$,
2. $\( \mathopen{}\left\langle{}\mathopen{}\left({x}_{1}, \dotsc, {x}_{n}\right)\mathclose{}, \mathopen{}\left({y}_{1}, \dotsc, {y}_{n}\right)\mathclose{}\right\rangle\mathclose{}= \sum_{i=1}^{n}{} \mathopen{}\left\langle{}{x}_{i}, {y}_{i}\right\rangle\mathclose{}_{i} \)$,
3. $\( \mathopen{}\left\lVert{}\mathopen{}\left({x}_{1}, \dotsb, {x}_{n}\right)\mathclose{}\right\rVert\mathclose{}= {\mathopen{}\left(\sum_{i=1}^{n}{} {\mathopen{}\left(\mathopen{}\left\lVert{}{x}_{i}\right\rVert\mathclose{}_{i}\right)\mathclose{}}^{2} \right)\mathclose{}}^{\frac{1}{2}} \)$.

It makes sense to use the same notation for the external direct sum $\( H= {H}_{1}\oplus \dotsb\oplus {H}_{n} \)$ and the (internal) direct sum $\( K= {H}_{1}\oplus \dotsb\oplus {H}_{n}\subseteq H \)$ since $\( H \)$ is isomorphic to $\( K \)$ in the following way: Let $\( {E}_{i} \)$ be the orthogonal projection of $\( K \)$ onto the subspace $\( {H}_{i} \)$. Note that $\( \sum_{i=1}^{n}{}{E}_{i}= I \)$. Now define $\( U : K \to H \)$ as $\( x\mapsto \mathopen{}\left({E}_{1}x, \dotsc, {E}_{n}x\right)\mathclose{} \)$. $\( U \)$ is clearly surjective. $\( U \)$ preserves norms since $\( {\mathopen{}\left\lVert{}U\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}= \sum_{i=1}^{n}{} {\mathopen{}\left\lVert{}{E}_{i}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2} = {\mathopen{}\left\lVert{}\mathopen{}\left(\sum_{i=1}^{n}{} {E}_{i} \right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2} \)$.

If $\( {V}_{1} \)$, …, $\( {V}_{n} \)$ are vector spaces, a function on $\( {V}_{1}\times \dotsb\times {V}_{n} \)$ is multilinear if it is linear on each $\( {V}_{i} \)$.

If $\( {V}_{1} \)$, …, $\( {V}_{n} \)$, $\( V \)$ are vector spaces, there exists a unique (up to isomorphism) vector space $\( W \)$ and a map $\( γ : {V}_{1}\times \dotsb\times {V}_{n} \to W \)$ with the property that if $\( φ : {V}_{1}\times \dotsb\times {V}_{n} \to V \)$ is a multilinear map, there exists a unique linear map $\( ψ : W \to V \)$ such that $\( φ= ψ\circ γ \)$.

The vector space $\( W \)$ and mapping $\( γ \)$ of Theorem I.84 is called the algebraic tensor product of $\( {V}_{1} \)$, …, $\( {V}_{n} \)$ and is written $\( {V}_{1}\otimes \dotsb\otimes {V}_{n} \)$. If $\( {x}_{i}\in {V}_{i} \)$, then $\( γ\mathopen{}\left( \mathopen{}\left({x}_{1}, \dotsc, {x}_{n}\right)\mathclose{}\right)\mathclose{} \)$ is written $\( {x}_{1}\otimes \dotsb\otimes {x}_{n} \)$.

Alternately, the algebraic tensor product may be thought of as the quotient space $\( \mathopen{}\left({V}_{1}\times \dotsb\times {V}_{n}\right)\mathclose{}/N \)$ where $\( N \)$ is the subspace spanned by elements of multilinear form.

If $\( V \)$ and $\( W \)$ are vector spaces with $\( {x}_{1} \)$ and $\( {x}_{2} \)$ in $\( V \)$ and with $\( {y}_{1} \)$ and $\( {y}_{2} \)$ in $\( W \)$, the following hold for $\( λ\in \mathbb{C} \)$:

1. $\( \mathopen{}\left(λ{x}_{1}\right)\mathclose{}\otimes {y}_{1}= {x}_{1}\otimes \mathopen{}\left(λ{y}_{1}\right)\mathclose{}= λ\mathopen{}\left({x}_{1}\otimes {y}_{1}\right)\mathclose{} \)$.
2. $\( \mathopen{}\left({x}_{1}+{x}_{2}\right)\mathclose{}\otimes {y}_{1}= {x}_{1}\otimes {y}_{1}+{x}_{2}\otimes {y}_{1} \)$.
3. $\( {x}_{1}\otimes \mathopen{}\left({y}_{1}+{y}_{2}\right)\mathclose{}= {x}_{1}\otimes {y}_{1}+{x}_{1}\otimes {y}_{2} \)$.

If $\( \mathopen{}\left(H, \mathopen{}\left\langle{}\cdot, \cdot\right\rangle\mathclose{}_{H}\right)\mathclose{} \)$ and $\( \mathopen{}\left(K, \mathopen{}\left\langle{}\cdot, \cdot\right\rangle\mathclose{}_{K}\right)\mathclose{} \)$ are Hilbert spaces, then $\( H\otimes K \)$ is an inner product space with inner product $$\[ \mathopen{}\left\langle{}{h}_{1}\otimes {k}_{1}, {h}_{2}\otimes {k}_{2}\right\rangle\mathclose{}= \mathopen{}\left\langle{}{h}_{1}, {h}_{2}\right\rangle\mathclose{}_{H}\mathopen{}\left\langle{}{k}_{1}, {k}_{2}\right\rangle\mathclose{}_{K} \]$$ and induced norm $$\[ \mathopen{}\left\lVert{}{h}_{1}\otimes {k}_{1}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}{h}_{1}\right\rVert\mathclose{}_{H}\mathopen{}\left\lVert{}{k}_{1}\right\rVert\mathclose{}_{K} \text{.} \]$$

If $\( H \)$ and $\( K \)$ are Hilbert spaces, the Hilbert space derived by completion of $\( H\otimes K \)$ with respect to the norm induced by the inner product of Theorem Theorem I.88 is called the Hilbert space tensor product of $\( H \)$ and $\( K \)$, and is denoted $\( H\mathbin{\hat{\otimes}} K \)$.

Sometimes the notation $\( H\odot K \)$ is used for the algebraic tensor product, and $\( H\otimes K \)$ for its completion.

Let $\( {μ}_{1} \)$ and $\( {μ}_{2} \)$ be measures on $\( \mathbb{R} \)$. The algebraic tensor product $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}, {μ}_{1}\right)\mathclose{}\otimes \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}, {μ}_{2}\right)\mathclose{} \)$ is composed of equivalence classes of functions, square integrable on $\( {\mathbb{R}}^{2} \)$ with respect to the measure $\( {μ}_{1}{μ}_{2} \)$. For $\( f \)$ and $\( g \)$ in $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}, {μ}_{1}\right)\mathclose{}\otimes \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}, {μ}_{2}\right)\mathclose{} \)$, $$\[ \mathopen{}\left\langle{}f, g\right\rangle\mathclose{}= \int {} \int {} \overline{ f\mathopen{}\left( x, y\right)\mathclose{} }g\mathopen{}\left( x, y\right)\mathclose{} \,\mathrm{d} {μ}_{1}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d} {μ}_{2}\mathopen{}\left( y\right)\mathclose{} \]$$ Since $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}, {μ}_{1}\right)\mathclose{}\otimes \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}, {μ}_{2}\right)\mathclose{} \)$ contains all step functions on $\( {\mathbb{R}}^{2} \)$, $$\[ \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}, {μ}_{1}\right)\mathclose{}\mathbin{\hat{\otimes}} \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}, {μ}_{2}\right)\mathclose{}\simeq \mathrm{L}^{\mathrm{2}}\mathopen{}\left( {\mathbb{R}}^{2}, {μ}_{1}\times {μ}_{2}\right)\mathclose{}\text{.} \]$$

If $\( {M}_{H} \)$ and $\( {M}_{K} \)$ are dense subspaces of $\( H \)$ and $\( K \)$ respectively, then $\( {M}_{H}\otimes {M}_{K} \)$ is dense in $\( H\mathbin{\hat{\otimes}} K \)$.

If $\( {E}_{H} \)$ and $\( {E}_{K} \)$ are orthonormal bases for $\( H \)$ and $\( K \)$ respectively, then $\( \mathopen{}\left\{\, {e}_{H}\otimes {e}_{K}\,\middle\vert\, {e}_{H}\in {E}_{H}, {e}_{K}\in {E}_{K}\,\right\}\mathclose{} \)$ is an orthonormal basis for $\( H\mathbin{\hat{\otimes}} K \)$. What does this say about the dimension of $\( H\mathbin{\hat{\otimes}} K \)$ ?

If $\( H \)$ and $\( K \)$ are non-trivial, then $\( H\mathbin{\hat{\otimes}} K \)$ is separable if and only if $\( H \)$ and $\( K \)$ are separable.

The inner product space defined in Theorem I.88 is complete if and only if $\( H \)$ or $\( K \)$ is finite dimensional.