[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 $x,y \in \mathbb{R}^n$.
[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]
[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]
