[proofplan] We expand the symmetric part $\operatorname{sym}(A)$ from its definition and use linearity of matrix multiplication. The only point that needs justification is that the two scalar quantities $x^\top A x$ and $x^\top A^\top x$ are equal. This follows by transposing the real $1 \times 1$ matrix $x^\top A x$. [/proofplan] [step:Expand the symmetric part inside the quadratic form] Fix $x \in \mathbb{R}^n$. By the definition of $\operatorname{sym}(A)$, \begin{align*} x^\top \operatorname{sym}(A)x = x^\top \left(\frac{1}{2}(A + A^\top)\right)x. \end{align*} Using scalar multiplication and distributivity of matrix multiplication over addition, we obtain \begin{align*} x^\top \operatorname{sym}(A)x = \frac{1}{2}x^\top A x + \frac{1}{2}x^\top A^\top x. \end{align*} [guided] Fix an arbitrary vector $x \in \mathbb{R}^n$. The goal is to compare the quadratic form defined by $A$ with the quadratic form defined by the symmetric part of $A$. Since the symmetric part is defined by \begin{align*} \operatorname{sym}(A) := \frac{1}{2}(A + A^\top), \end{align*} substituting this definition into the expression $x^\top \operatorname{sym}(A)x$ gives \begin{align*} x^\top \operatorname{sym}(A)x = x^\top \left(\frac{1}{2}(A + A^\top)\right)x. \end{align*} Matrix multiplication is compatible with scalar multiplication, so the factor $\frac{1}{2}$ may be pulled outside. Matrix multiplication is also distributive over matrix addition, so the product with $A + A^\top$ splits into the sum of the products with $A$ and with $A^\top$. Therefore \begin{align*} x^\top \operatorname{sym}(A)x = \frac{1}{2}x^\top A x + \frac{1}{2}x^\top A^\top x. \end{align*} This reduces the theorem to showing that the two real scalars $x^\top A x$ and $x^\top A^\top x$ are equal. [/guided] [/step] [step:Identify the transpose term with the original quadratic form] The product $x^\top A x$ is a $1 \times 1$ real matrix, hence it is equal to its transpose. Using the transpose rule for products, \begin{align*} (x^\top A x)^\top = x^\top A^\top x. \end{align*} Thus \begin{align*} x^\top A^\top x = x^\top A x. \end{align*} [/step] [step:Conclude that the skew contribution cancels] Substituting $x^\top A^\top x = x^\top A x$ into the expansion from the first step gives \begin{align*} x^\top \operatorname{sym}(A)x = \frac{1}{2}x^\top A x + \frac{1}{2}x^\top A x. \end{align*} Therefore \begin{align*} x^\top \operatorname{sym}(A)x = x^\top A x. \end{align*} Since $x \in \mathbb{R}^n$ was arbitrary, the identity holds for every $x \in \mathbb{R}^n$. [/step]