I want show that
Let $T: \Bbb R^n \to \Bbb R^n$ be a symmetric linear transformation, i.e. $T(\mathbf{x}) \cdot \mathbf{y} = \mathbf{x} \cdot T(\mathbf{y})$ for all $\mathbf{x},\mathbf{y} \in \Bbb R^n$. If $B$ is any orthonormal basis of $\Bbb R^n$, then the matrix $[T]_B$ is symmetric.
Why is this only true for an orthonormal basis? I tried to prove it and showed it for any basis and so I must have made a mistake. Where has it gone wrong? I found those two questions here and here but they didn't explain that.
Let $B=\{v_1,\ldots,v_n\}$ be a basis for $\Bbb R^n$. We know that $[T]_B [\mathbf{x}]_B \cdot [\mathbf{y}]_B = [\mathbf{x}]_B \cdot [T]_B [\mathbf{y}]_B$ is true for all $\mathbf{x},\mathbf{y} \in \Bbb R^n$. So it must especially be true for all $\mathbf{v}_i$ which means $[T]_B [\mathbf{v}_i]_B \cdot [\mathbf{v}_j]_B = [\mathbf{v}_i]_B \cdot [T]_B [\mathbf{v}_j]_B$ for every $1 \le i,j \le n$. Now $[\mathbf{v}_j]_B$ has a $1$ in the $i$-th slot and zeros everywhere else. Therefore for the LHS we get from the matrix multiplication of $[T]_B [\mathbf{v}_i]_B$ the $i$-th row of $[T]_B$ and the scalar product with $[\mathbf{v}_j]_B$ gives the $j$-th entry from this $i$-th row. Analogously for the RHS. Thus we get $\left([T]_B\right)_{ji}=\left([T]_B \right)_{ij}$ which means $[T]_B = [T]_B^T$ and so $[T]_B$ is symmetric.