[proofplan] We prove the two implications separately. If $A$ is symmetric, then $A^\top=A$, and the defining matrix formula for the Euclidean [inner product](/page/Inner%20Product) gives the desired identity by moving the transpose from $Ax$ onto the matrix. Conversely, assuming the inner-product identity for all vectors, we test it on pairs of standard basis vectors $e_i,e_j$; these tests isolate the entries $A_{ji}$ and $A_{ij}$, forcing equality of all transposed entries and hence symmetry. [/proofplan]

[step:Use transpose algebra to prove the inner-product identity from symmetry]Assume that $A$ is symmetric. By definition, this means $A^\top=A$. Let $x,y \in \mathbb{R}^n$ be arbitrary. The Euclidean inner product on $\mathbb{R}^n$ is given by $\langle u,v\rangle=u^\top v$ for $u,v \in \mathbb{R}^n$. Therefore, \begin{align*} \langle Ax,y\rangle=(Ax)^\top y. \end{align*} Using the transpose product rule $(Ax)^\top=x^\top A^\top$, we obtain \begin{align*} (Ax)^\top y=x^\top A^\top y. \end{align*} Since $A^\top=A$, this becomes \begin{align*} x^\top A^\top y=x^\top Ay. \end{align*} Finally, by the definition of the Euclidean inner product again, \begin{align*} x^\top Ay=\langle x,Ay\rangle. \end{align*} Combining these equalities gives $\langle Ax,y\rangle=\langle x,Ay\rangle$ for all $x,y \in \mathbb{R}^n$.[/step]

[guided]Assume that $A$ is symmetric. The meaning of symmetry for a real square matrix is exactly the identity $A^\top=A$. We want to show that $A$ can be moved from the first slot of the Euclidean inner product to the second slot without changing the value. Let $x,y \in \mathbb{R}^n$ be arbitrary. Since the Euclidean inner product is defined by $\langle u,v\rangle=u^\top v$, applied to $u=Ax$ and $v=y$ it gives \begin{align*} \langle Ax,y\rangle=(Ax)^\top y. \end{align*} The transpose product rule gives $(Ax)^\top=x^\top A^\top$, so \begin{align*} (Ax)^\top y=x^\top A^\top y. \end{align*} Now the symmetry hypothesis is used: because $A^\top=A$, we may replace $A^\top$ by $A$ and obtain \begin{align*} x^\top A^\top y=x^\top Ay. \end{align*} But $x^\top Ay$ is precisely the Euclidean inner product of $x$ with $Ay$: \begin{align*} x^\top Ay=\langle x,Ay\rangle. \end{align*} Thus $\langle Ax,y\rangle=\langle x,Ay\rangle$. Since $x$ and $y$ were arbitrary vectors in $\mathbb{R}^n$, the identity holds for all $x,y \in \mathbb{R}^n$.[/guided]

[step:Test the inner-product identity on standard basis vectors to recover symmetry] Conversely, assume that \begin{align*} \langle Ax,y\rangle=\langle x,Ay\rangle \end{align*} for all $x,y \in \mathbb{R}^n$. For each $i \in \{1,\dots,n\}$, let $e_i \in \mathbb{R}^n$ denote the $i$-th standard basis vector, whose $k$-th coordinate is $1$ if $k=i$ and $0$ otherwise. Fix arbitrary indices $i,j \in \{1,\dots,n\}$. Applying the assumed identity with $x=e_i$ and $y=e_j$ gives \begin{align*} \langle Ae_i,e_j\rangle=\langle e_i,Ae_j\rangle. \end{align*} The vector $Ae_i$ is the $i$-th column of $A$, so its $j$-th coordinate is $A_{ji}$. Hence \begin{align*} \langle Ae_i,e_j\rangle=A_{ji}. \end{align*} Similarly, the vector $Ae_j$ is the $j$-th column of $A$, so its $i$-th coordinate is $A_{ij}$. Hence \begin{align*} \langle e_i,Ae_j\rangle=A_{ij}. \end{align*} Therefore $A_{ji}=A_{ij}$. Since $i,j \in \{1,\dots,n\}$ were arbitrary, all transposed entries agree: \begin{align*} A_{ij}=A_{ji} \end{align*} for every $i,j \in \{1,\dots,n\}$. Thus $A^\top=A$, so $A$ is symmetric. [/step]

[step:Combine the two implications] The first step proves that every [symmetric matrix](/page/Symmetric%20Matrix) satisfies the stated Euclidean inner-product identity. The second step proves that any matrix satisfying the identity for all $x,y \in \mathbb{R}^n$ has entries $A_{ij}=A_{ji}$ for all $i,j \in \{1,\dots,n\}$, hence is symmetric. Therefore $A$ is symmetric if and only if \begin{align*} \langle Ax,y\rangle=\langle x,Ay\rangle \end{align*} for all $x,y \in \mathbb{R}^n$. [/step]