<s>[INST] Can a symmetric bilinear map $V\times V \to W$ increase subspace dimension? Let $V$ and $W$ be two finite-dimensional real vector spaces, and let $S$ be a subspace of $V$. By the rank-nullity theorem, if $T$ is a linear map $V \to W$, then $\dim T(S) \leq \dim S$.
Now, let $B$ denote a symmetric bilinear map $V \times V \to W$. I would like to know if a similar statement holds for $B$.
More precisely, consider the subspace defined by the span of the image of $B\lvert_{S\times S}$:
$$
B(S)=\mathrm{span}\,\{B(x,y) \mid x,y \in S \}.
$$
Is it true that $\dim B(S) \leq \dim S$?
 [/INST] Take $V = \mathbb{R}^2_{\textrm{col}}$ and consider the map $B \colon V \times V \rightarrow M_2(\mathbb{R})$ given by $B(x,y) = \frac{1}{2} \left( x \cdot y^T + y \cdot x^T \right)$. Then
$$ B(e_1,e_1) = e_1 \cdot e_1^T = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \\
B(e_1,e_2) = \frac{1}{2} \left( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \end{pmatrix} \right) = \frac{1}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\\
B(e_2,e_2) = e_2 \cdot e_2^T = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.   $$
This means that $B(V)$ is the subspace of $2 \times 2$ symmetric matrices which is three dimensional while $V$ has dimension two.
</s>