We consider the different flavours of inner product in turn.

Antisymmetric Clearly, for a \(1\)-dimensional vector space the only possible antisymmetric inner product is the zero inner product. In the \(2\)-dimensional case, suppose first that the skew-symmetric inner product is degenerate. Then there exists, in our \(2\)-dimensional space \(V\), a non-zero vector \(v\) such that \((u,v)=0\) \(\forall u\in V\) and we can extend \(v\) to a basis \(\{v,v’\}\) of \(V\). Now consider the inner product of two arbitrary elements, \(av+bv’\) and \(cv+dv’\), where \(a\), \(b\), \(c\) and \(d\) are elements of the base field (\(\RR\) or \(\CC\)). We have
\begin{equation*}
(av+bv’,cv+dv’)=ac(v,v)+ad(v,v’)+bc(v’,v)+bd(v’,v’)=0,
\end{equation*}
so the only degenerate antisymmetric inner product on a \(2\)-dimensional vector space is the zero inner product. So consider the non-degenerate case. There must exist two vectors, \(v_1\) and \(v_2\), say, such that \((v_1,v_2)\neq0\). In particular this means they are linearly independent and so we can take them to be a basis of \(V\). If \((v_1,v_2)=a\) then the map \(f:V\mapto K^2\) defined by \(f(v_1)=a\mathbf{e}_1\), \(f(v_2)=\mathbf{e}_2\) is clearly an isometry between \(V\) and the symplectic space \(K^2\) of Example. That is, any symplectic geometry on a 2-dimensional vector space is isometric to either the trivial, zero case, or that of Example.

Symmetric Consider first a \(1\)-dimensional vector space, \(V\), over \(\RR\). If \(v\in V\) is non-zero but \((v,v)=0\) then we have the trivial case. So suppose \((v,v)=a\) for some \(a\in\RR\). Either \(a>0\), in which case the inner product is positive definite and we have isometry with the inner product space \(\RR\) equipped with the inner product given by simple multiplication, \((x,y)=xy\) for \(x,y\in\RR\), or \(a<0\), in which case the inner product is negative definite and we have isometry with \(\RR\) equipped with the inner product, \((x,y)=-xy\). As already observed, these two cases are not isometric. If, however, \(V\) is over \(\CC\) then any non-degenerate inner product space is simply isometric to \(\CC\) with the inner product \((x,y)=xy\). Hermitian Any non-trivial \(1\)-dimensional Hermitian inner product space \(V\) must be such that \((v,v)\neq0\) for some \(v\in V\). In this case \((v,v)=a\) with \(a\) some non zero real number. Similar to the real symmetric case we have two cases, positive or negative definite, each clearly isometric to \(\CC\) equipped respectively with the Hermitian inner product \((x,y)=x^*y\) or \((x,y)=-x^*y\).