Proof of matrix-vector multiplication commutativeness for a symmetric matrix

For vectors \(\mathbf{a}, \mathbf{b} \in \mathbf{R}^n\) and matrix \(K \in \mathbf{R}^{n \times n}\), we may write

\[\begin{align} \mathbf{a}^T K \mathbf{b} &= a^T \begin{bmatrix} K_{11}b_1 + K_{12}b_2+\cdots+K_{1n}b_n, \hspace{10px} \cdots, %%%%K_{21}b_1 + K_{22}b2 + \cdots+K_{2n}b_n, \hspace{10px} K_{n1}b_1 + K_{n2}b_2 + \cdots+K_{nn}b_n \end{bmatrix}\\ &= K_{11}(a_1b_1)+ K_{12}(a_1b_2)+\cdots+K_{1n}(a_1b_n)+\cdots\\&+K_{n1}(a_nb_1)+K_{n2}(a_nb_2)+K_{nn}(a_nb_n), \end{align}\]

and similarly

\[\begin{align} \mathbf{b}^T K \mathbf{a} &= b^T \begin{bmatrix} K_{11}a_1 + K_{12}a_2+\cdots+K_{1n}a_n, \hspace{10px} \cdots, %%%%K_{21}b_1 + K_{22}b2 + \cdots+K_{2n}b_n, \hspace{10px} K_{n1}a_1 + K_{n2}a_2 + \cdots+K_{nn}a_n \end{bmatrix}\\ &= K_{11}(b_1a_1)+K_{12}(b_1a_2)+\cdots+K_{1n}(b_na_1)+\cdots\\&+K_{n1}(b_na_1)+K_{n2}(b_na_2)+K_{nn}(b_na_n). \end{align}\]

We can see a pattern emerging here, where all factors of the diagonal of \(K\) are equal, such as \(K_{11}(a_1b_1)=K_{11}(b_1a_1)\). The differences are in the non-diagonal elements, such as the factor of \(K_{12}\) being \((a_1b_2)\) for \(\mathbf{a}^TK\mathbf{b}\) but \((a_2b_1)\) for \(\mathbf{b}^TK\mathbf{a}\). Intuitively, if \(K_{12}=K_{21}\), the equality holds. It can be verified by developing the above linear systems that \(\mathbf{a}^TK\mathbf{b} = \mathbf{b}^TK\mathbf{a}\) if and only if \(K_{ij} = K_{ji}\), or in other words if \(K^T=K\).