Definition Two vectors \(v,w\in V\) are said to be orthogonal if \((v,w)=0\). If \(U\) is a subspace of \(V\) then the orthogonal complement of \(U\), denoted \(U^\perp\), is defined as
\begin{equation}
U^\perp=\{v\in V\mid(u,v)=0\;\forall u\in U\}.
\end{equation}

It is clear that \(U^\perp\) is also a subspace of \(V\) and that if the restriction of the inner product to \(U\) is non-singular, then \(U\cap U^\perp=\{0\}\). In fact, in this case we have the following result.

Theorem If \(U\) is a subspace of an inner product space \(V\) such that the restriction of the inner product to \(U\) is non-degenerate, then \(V=U\oplus U^\perp\).

Proof Since we already know that \(U\cap U^\perp=\{0\}\), we need only demonstrate that any \(v\in V\) can be written as \(v=u+v’\) with \(u\in U\) and \(v’\in U^\perp\). To this end suppose \(e_1,\dots,e_r\) is a basis of \(U\). Then we must find numbers \(c^i\) and some \(v’\in U^\perp\) such that
\begin{equation}
v=c^1e_1+\dots+c^re_r+v’.\label{equ:orthog comp intermediate}
\end{equation}
Now define the matrix \(\mathbf{M}\) through the matrix elements, \(M_{ij}=(e_i,e_j)\). Then taking succesive inner products of \eqref{equ:orthog comp intermediate} with the basis elements of \(U\), we get the system of equations,
\begin{eqnarray*}
(e_1,v)&=&M_{11}c_1+\dots+M_{rr}c_r\\
\vdots\quad&\vdots&\qquad\quad\vdots\\
(e_r,v)&=&M_{r1}c_1+\dots+M_{rr}c_r,
\end{eqnarray*}
and since the restriction of the inner product to \(U\) is non-degenerate, \(\mathbf{M}\) is non-singular. Thus there is a unique solution for the \(c^i\), so any \(v\in V\) can be expressed in the form, \eqref{equ:orthog comp intermediate}, and the result follows.\(\blacksquare\)

Remark Recall that a direct sum decomposition, \(V=U_1\oplus U_2\), determines projectors, \(P_1,P_2\in\mathcal{L}(V)\), \(P_i^2=P_i\), such that, \(P_1+P_2=\id_V\) and \(P_iP_j=0\) when \(i\neq j\), and, \(\img P_i=U_i\), \(\ker P_1=U_2\) and \(\ker P_2=U_1\). In the context of inner product spaces orthogonal projections are natural. These are the projections corresponding to an orthogonal direct sum decomposition, such as \(V=U\oplus U^\perp\), that is, projections whose image and kernel are orthogonal.

A non-zero vector \(v\) of an inner product space \(V\) is said to be a null vector if \((v,v)=0\). All vectors are null in symplectic geometries. In the case of orthogonal or Hermitian geometries, aside from the trivial case of a zero inner product, all vectors cannot be null. Indeed, suppose on the contrary, that every \(v\in V\) was such that \((v,v)=0\). Then for every pair of vectors \(u,v\in V\), we have, in the case of a symmetric inner product,
\begin{equation*}
0=(u+v,u+v)=(u,u)+(v,v)+2(u,v)=2(u,v),
\end{equation*}
so \((u,v)=0\) implying the inner product is zero. In the case of an Hermitian inner product,
\begin{equation*}
0=(u+v,u+v)=(u,u)+(v,v)+2\Real(u,v)=2\Real(u,v),
\end{equation*}
and
\begin{equation*}
0=(u+iv,u+iv)=(u,u)+(v,v)+2i\Imag(u,v)=2i\Imag(u,v),
\end{equation*}
so also in this case, \((u,v)=0\), contradicting our assumption that inner product is non-zero.

Theorem Any finite dimensional inner product space, \(V\), over \(\RR\) or \(\CC\), can be decomposed into a direct sum \(V=V_1\oplus\dots\oplus V_r\) where the subspaces \(V_i\) are pairwise orthogonal. In the case of symmetric or Hermitian inner products they are \(1\)-dimensional. In the case of an anti-symmetric inner product the \(V_i\) may be either \(1\)-dimensional, in which case the restriction of the inner product to \(V_i\) is degenerate, or \(2\)-dimensional, in which case the restriction is non-degenerate.

Proof The proof is by induction on the dimension of \(V\). The case of \(\dim V=1\) is trivial so consider \(\dim V\geq 2\). We assume the inner product is not zero, since in this trivial case there is nothing to prove. In the case of symmetric or Hermitian inner products, as already observed, we can choose a non-null vector \(u\) from \(V\). If \(U=\Span(u)\) then the restriction of the inner product of \(V\) to \(U\) is certainly non-degenerate so we have \(V=U\oplus U^\perp\) by the previous result. Thus, if \(V\) is \(n\) dimensional then \(U^\perp\) is \(n-1\) dimensional and by induction we can therefore assume that \(U^\perp\) has the desired decomposition and the result follows. In the case of an anti-symmetric inner product, there must exist two vectors, \(v_1\) and \(v_2\) say, such that \((v_1,v_2)\neq0\). Call the subspace spanned by these vectors \(U\) then the restriction of the inner product to \(U\) is non-degenerate and the result follows as before.\(\blacksquare\)

Given this orthogonal decomposition we can use what we already know about the classification of low dimensional inner product spaces to complete the classification in general. To that end, we consider two \(n\)-dimensional vector spaces, \(V\) and \(V’\), with inner products, \((\cdot,\cdot)\) and \((\cdot,\cdot)’\), and orthogonal decompositions \(\oplus_{i=1}^rV_i\) and \(\oplus_{i=1}^{r’}{V_i}’\) respectively.

Define the subspace, \(V_0=\oplus_{i=1}^{r_0}V_i\), the sum of the degenerate subspaces of the orthogonal decomposition, and \(V^\perp=\{v\in V\mid(v’,v)=0\;\forall v’\in V\}\), sometimes called the radical of the inner product \((\cdot,\cdot)\). Clearly \(V_0\subseteq V^\perp\) and conversely, by virtue of the decomposition, we know that any \(v\in V^\perp\) can be written uniquely as a sum \(v=\sum_{i=1}^nv_i\) with each \(v_i\in V_i\). Suppose it was the case that \(v_k\neq0\) for some \(k>r_0\). Then, in the case of symmetric or Hermitian inner products we’d have \((v_k,v_k)\neq0\) and in the anti-symmetric case there’d be some vector \({v’}_k\in V_k\) such that \(({v’}_k,v_k)\neq0\) so in either case we contradict \(v\in V^\perp\) and conclude that \(V_0=V^\perp\). But we also have that if \(\{e_i\}\) is a basis of \(V\) in which the Gram matrix is \(\mathbf{G}\) then \(V^\perp\) consists of those \(v=v^ie_i\in V\) such that \((e_j,v)=0\) for each \(j=1,\dots,n\), that is, such that, \(\sum_{i=1}^nG_{ji}v^i=0\), for each \(j=1,\dots,n\). But this is just the condition for \(v\in\ker L_\mathbf{G}\), so \(V^\perp=\ker L_\mathbf{G}\) and \(\dim V^\perp=n-\rank\mathbf{G}\). Thus, we may conclude that \(r_0=\dim V^\perp=n-\rank\mathbf{G}\). Moreover, since the Gram matrices of isometric inner product spaces have the same rank, if our two inner product spaces \(V\) and \(V’\) are isometric we may further conclude that \(r_0={r_0}’\) where \({r_0}’\) is the number of degenerate subspaces in the orthogonal decomposition of \(V’\).

Now let us consider the case of \(V\) and \(V’\) having anti-symmetric inner products. We know that aside from \(1\)-dimensional degenerate subspaces all the remaining subspaces in their respective orthogonal decompositions are \(2\)-dimensional non-degenerate and that any two such spaces are isometric. Thus, if \(V\) and \(V’\) have \(r_0={r_0}’\) then they must both have \(r_0\) degenerate \(1\)-dimensional subspaces and \((n-r_0)/2\) non-degenerate \(2\)-dimensional subspaces in their respective decompositions, which we may order such that in both cases non-degenerate precede degenerate subspaces. Therefore they must be isometric, since we can construct an isometry \(f:V\mapto V’\) as the direct sum \(f=\oplus_{i=1}^rf_i\) of isometries \(f_i:V_i\mapto {V_i}’\) which we know must exist from the discussion in Low Dimensional Inner Product Spaces. Conversely, if \(V\) and \(V’\) are isometric we know that \(r_0={r_0}’\). Thus, we conclude that two vector spaces equipped with antisymmetric inner products are isometric if and only if the vector spaces and their respective radicals have the same dimension. It should be clear that precisely the same statement can be made for two complex vector spaces with symmetric inner products.

In the orthogonal decompositions of real vector spaces equipped with symmetric, or complex vector spaces equipped with Hermitian, inner products, aside from the degenerate subspaces, there are in each case two possibilities for the remaining \(1\)-dimensional subspaces. They may be either positive or negative definite. Denote by \(r_+\) and \(r_-\) respectively the number of positive and negative definite subspaces of \(V\). If \(V_+\) and \(V_-\) are the respective direct sums of these subspaces then it is clear that \(V_+\) and \(V_-\) are respectively positive and negative definite, that \(r_+=\dim V_+\) and \(r_-=\dim V_-\) and that we can write \(V=V_+\oplus V_-\oplus V_0\). The triple \((r_+,r_-,r_0)\) is called the signature of the inner product. Define the same primed quantities, \({r_+}’\), \({V’}_+\), \({r_-}’\) and \({V’}_-\) for \(V’\), whose decomposition can then be written as, \(V’={V’}_+\oplus {V’}_-\oplus {V’}_0\). Now suppose \(V\) and \(V’\) are isometric. We know that they must have the same dimension and that \(r_0={r_0}’\). If \(f:V\mapto V’\) is an isometry, then for any \(v\in V\), by virtue of the decomposition of \(V’\), we can write \(f(v)=f(v)_++f(v)_-+f(v)_0\) where \(f(v)_+\in {V’}_+\), \(f(v)_-\in {V’}_-\) and \(f(v)_0\in {V’}_0\). We consider the restriction \(f|_{V_+}\) of \(f\) to \(V_+\) and note that if \({P’}_+:V’\mapto V’\) is the projection operator onto the subspace \({V’}_+\), then \({P’}_+\circ f|_{V_+}:V_+\mapto {V’}_+\) is linear. Now suppose \(r_+>{r_+}’\), then there must exist some \(v\in V_+\) such that \({P’}_+\circ f|_{V_+}(v)=0\), so that for this \(v\), \(f(v)_+=0\) and we have \(f(v)=f(v)_-+f(v)_0\). But notice that then \((v,v)=(f(v),f(v))’=(f(v)_-,f(v)_-)’<0\) contradicting the fact that \(v\in V_+\). Similarly, if \(r_+<{r_+}'\), then we must have \(r_->{r_-}’\), and we would again arrive at a contradiction by considering the restriction \(f|_{V_-}\). So we conclude that isometry of \(V\) and \(V’\) implies \(r_+={r_+}’\), \(r_-={r_-}’\) and \(r_0={r_0}’\), that is the have the same signature. Conversely, if two vector spaces \(V\) and \(V’\) have the same signature, then their orthogonal decompositions can be appropriately ordered such that an isometry, \(f:V\mapto V’\), can be constructed as the direct sum, \(f=\oplus f_i\), of isometries, \(f_i:V_i\mapto V_i\), which we know must exist from the discussion in Low Dimensional Inner Product Spaces. Thus, we conclude that two real vector spaces equipped with symmetric inner products or two complex vector spaces equipped with Hermitian inner products are isometric if and only if they share the same signature.

Let us summarise the above discussion in the following

Theorem Symplectic spaces and complex orthogonal spaces are characterised up to isometry by the pair of integers \((n,r_0)\) where \(n\) is the dimension of the space and \(r_0\) is the dimension of the radical of the inner product. Real orthogonal spaces and complex Hermitian spaces are characterised up to isometry by their signature, \((r_+,r_-,r_0)\), where \(r_+\) and \(r_-\) are the dimensions of the subspaces upon which the restriction of the inner product is respectively positive and negative definite.

Theorem and Theorem tell us that for orthogonal and Hermitian spaces we can always find an orthogonal basis. Such a basis is particularly useful when the inner product is non-degenerate. In this case, any vector \(v\) may be expressed as \(v=v^je_j\), but then \((e_i,v)=(e_i,v^je_j)\) so
\begin{equation}
v=\sum_{i=1}^n\frac{(e_i,v)}{(e_i,e_i)}e_i.
\end{equation}
A vector \(v\) is said to be normalised if \((v,v)=\pm1\), and a set of normalised vectors \(\{e_i\}\) is said to be orthonormal if they are orthogonal with one another, that is, \((e_i,e_j)=0\) whenever \(i\neq j\) and \((e_i,e_i)=\pm1\). So given any orthogonal basis of a non-degenerate inner product space we can always choose an orthonormal basis. If the inner product is positive definite then with respect to an orthonormal basis any vector the convenient decomposition,
\begin{equation}
v=\sum_{i=1}^n(e_i,v)e_i.
\end{equation}

It should be clear that for real orthogonal and complex Hermitian spaces we can always find a basis in which the Gram matrix has the form,
\begin{equation}
\mathbf{G}=\begin{pmatrix}
\mathbf{I}_{r_+}&\mathbf{0}&\mathbf{0}\\
\mathbf{0}&-\mathbf{I}_{r_-}&\mathbf{0}\\
\mathbf{0}&\mathbf{0}&\mathbf{0}
\end{pmatrix},
\end{equation}
and for complex orthogonal spaces, a basis in which the Gram matrix has the form,
\begin{equation}
\mathbf{G}=\begin{pmatrix}
\mathbf{I}_{n-r_0}&\mathbf{0}\\
\mathbf{0}&\mathbf{0}
\end{pmatrix}.
\end{equation}

Clearly there’s no such thing as an orthogonal basis for a symplectic space. However, Theorem and Theorem do make it clear that we can always choose a basis, \(\{e_1,f_1,\dots,e_{(n-r_0)/2},f_{(n-r_0)/2},e_{(n-r_0)/2+1},\dots,e_{(n+r_0)/2}\}\), such that, \((e_i,f_i)=-(f_i,e_i)=1\) for \(i=1,\dots,(n-r_0)/2\), are the only non-zero inner products of basis elements. Reordering, we obtain the symplectic basis,
\begin{equation}
\{e_1,\dots,e_{(n-r_0)/2},f_1,\dots,f_{(n-r_0)/2},e_{(n-r_0)/2+1},\dots,e_{(n+r_0)/2}\},
\end{equation}
in terms of which, the Gram matrix has the form,
\begin{equation}
\mathbf{G}=\begin{pmatrix}
\mathbf{0}&\mathbf{I}_{(n-r_0)/2}&\mathbf{0}\\
-\mathbf{I}_{(n-r_0)/2}&\mathbf{0}&\mathbf{0}\\
\mathbf{0}&\mathbf{0}&\mathbf{0}
\end{pmatrix}
\end{equation}

Hermitian, real orthogonal and real symplectic geometries arise quite naturally together, in the following way. Suppose we have a complex vector space \(V\) on which is defined the Hermitian inner product \((\cdot,\cdot):V\times V\mapto\CC\). We consider its realification, \(V_\RR\), on which we define two inner products, \(g(v,w)=\Real(v,w)\) and \(\omega(v,w)=\Imag(v,w)\). It is clear that \(g\) is symmetric and \(\omega\) is antisymmetric and that therefore, \(g\) is positive definite if and only if \((\cdot,\cdot)\) is positive definite. We also have the following relations,
\begin{eqnarray}
g(v,w)=g(iv,iw)=\omega(v,iw)=-\omega(iv,w)\label{orthsympl1}\\
\omega(v,w)=\omega(iv,iw)=-g(v,iw)=g(iv,w).\label{orthsympl2}
\end{eqnarray}
Conversely, if on \(V_\RR\), there are defined inner products, \(g\) and \(\omega\), respectively symmetric and antisymmetric, which satisfy the relations \eqref{orthsympl1} and \eqref{orthsympl2}, then the inner product on \(V\) defined as \((v,w)=g(v,w)+i\omega(v,w)\) is Hermitian.

Consider, in particular, \(\CC^n\) with the standard (orthonormal) basis \(\{\mathbf{e}_i\}\) and Hermitian inner product,
\begin{equation*}
(\mathbf{v},\mathbf{w})=\sum_{i=1}^n{v_i}^*w_i.
\end{equation*}
It’s realification is \(\RR^{2n}\) with basis \(\{\mathbf{e}_1,\dots,\mathbf{e}_n,i\mathbf{e}_1,\dots,i\mathbf{e}_n\}\) which is orthonormal with respect to \(g\) and symplectic with respect to \(\omega\).

The Hermitian inner product, or more precisely its absolute value, measures the extent to which two vectors are parallel or linearly dependent over \(\CC\) while \(g\) measures this over \(\RR\). Thus, \(\omega\) measures the extent to which the linear dependence of two vectors is due to extending the base field from \(\RR\) from \(\CC\).

The upshot of this strong relationship, particularly between complex Hermitian and real orthogonal inner products, is that we can develop their theory largely in parallel.