[proofplan] Both directions follow from the relationship $M_{ij} = \phi(e_i, e_j)$. The forward direction is immediate from the symmetry condition. The backward direction expands $\phi(v, w)$ and $\phi(w, v)$ in coordinates and uses the fact that a $1 \times 1$ matrix equals its transpose to identify the two expressions. [/proofplan] [step:Show symmetric $\phi$ implies symmetric $M$] If $\phi$ is symmetric, then for all $i, j \in \{1, \dots, n\}$: \begin{align*} M_{ij} = \phi(e_i, e_j) = \phi(e_j, e_i) = M_{ji}. \end{align*} So $M = M^\top$. [/step] [step:Show symmetric $M$ implies symmetric $\phi$ by expanding in coordinates] Suppose $M = M^\top$. For $v = \sum_i v_i e_i$ and $w = \sum_j w_j e_j$, bilinearity gives: \begin{align*} \phi(v, w) = \sum_{i,j} v_i w_j M_{ij} = v^\top M w. \end{align*} Similarly, $\phi(w, v) = w^\top M v$. Since $w^\top M v$ is a $1 \times 1$ scalar, it equals its own transpose: \begin{align*} w^\top M v = (w^\top M v)^\top = v^\top M^\top w = v^\top M w, \end{align*} where the last equality uses $M = M^\top$. Therefore $\phi(v, w) = \phi(w, v)$ for all $v, w \in V$. [guided] Assume $M = M^\top$. We want to show $\phi(v, w) = \phi(w, v)$ for all $v, w \in V$. Expanding in the basis $(e_1, \dots, e_n)$ with $v = \sum_i v_i e_i$ and $w = \sum_j w_j e_j$: \begin{align*} \phi(v, w) = \sum_{i=1}^n \sum_{j=1}^n v_i\, w_j\, \phi(e_i, e_j) = \sum_{i,j} v_i\, w_j\, M_{ij}. \end{align*} In matrix notation this is $v^\top M w$, where we treat $v$ and $w$ as column vectors of coordinates. The expression $w^\top M v$ is a $1 \times 1$ matrix, so it equals its transpose: \begin{align*} \phi(w, v) = w^\top M v = (w^\top M v)^\top = v^\top M^\top w. \end{align*} Since $M^\top = M$ by hypothesis, this gives $\phi(w, v) = v^\top M w = \phi(v, w)$. Note that symmetry of $M$ is preserved under congruence: if $B = P^\top M P$ is the matrix of $\phi$ in another basis, then $B^\top = (P^\top M P)^\top = P^\top M^\top P = P^\top M P = B$. So the property of having a symmetric matrix does not depend on the choice of basis, as expected since symmetry of $\phi$ is an intrinsic property. [/guided] [/step]