Here $V$ be a vsp over a field $K$ > [!NOTE] > Suppose that $2 \neq 0$ in $K$. Let $\tau:V\to K$ be a [[Bilinearity|bilinear form]]. If $\tau = \tau_{1} + \tau_{2}$where $\tau_{1}$ is symmetric and $\tau_{2}$ is antisymmetric then $\tau_{1}(v,w) = \frac{1}{2} (\tau(v,w)+\tau(w,v))$ and $\tau_{2}(v,w)=\frac{1}{2}(\tau(v,w) - \tau(w,v))$ ###### Proof It's clear that $\tau_1$ is symmetric and $\tau_2$ is antisymmetric. Moreover, given any other such expression $\tau=\tau_1^{\prime}+\tau_2^{\prime}$, we have $ \begin{aligned} \tau_1(\mathbf{v}, \mathbf{w}) & =\frac{\tau_1^{\prime}(\mathbf{v}, \mathbf{w})+\tau_1^{\prime}(\mathbf{w}, \mathbf{v})+\tau_2^{\prime}(\mathbf{v}, \mathbf{w})+\tau_2^{\prime}(\mathbf{w}, \mathbf{v})}{2} \\ & =\frac{\tau_1^{\prime}(\mathbf{v}, \mathbf{w})+\tau_1^{\prime}(\mathbf{v}, \mathbf{w})+\tau_2^{\prime}(\mathbf{v}, \mathbf{w})-\tau_2^{\prime}(\mathbf{v}, \mathbf{w})}{2} \end{aligned} $ from the symmetry and antisymmetry of $\tau_1^{\prime}$ and $\tau_2^{\prime}$. The last two terms cancel each other and we just have $ =\frac{2 \tau_1^{\prime}(\mathbf{v}, \mathbf{w})}{2}=\tau_1^{\prime}(\mathbf{v}, \mathbf{w}) $ So $\tau_1=\tau_1^{\prime}$, and so $\tau_2=\tau-\tau_1=\tau-\tau_1^{\prime}=\tau_2^{\prime}$, so the decomposition is unique.