If the inner product is symmetric iff $A=A^{T}$ In class we defined the following inner product:
Let $A\in M_n(\mathbb{R})$, and let $\langle \cdot,\cdot \rangle :\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$ be defined as:
$$\langle x,y\rangle=x^TAy.$$
Now, I need to prove that the inner product is symmetric iff $A=A^{T}$.
Proof:
First direction:
If $A=A^{T}$, then $\langle Ax,y \rangle = \langle x,Ay \rangle$
$$\langle Ax,y \rangle =(Ax)^TAy=x^TA^TAy=x^TAAy=\langle x,Ay\rangle$$
The other direction:
Suppose that $\langle Ax,y \rangle = \langle x,Ay \rangle$, we want to show that $A=A^{T}$.
Here I chose to look at $x=e_i$ and $y=e_j$, standard basis vectors of $\mathbb{R}^n$.
I asked myself, what is $Ax$?
$Ax$ is column $i$ of $A$, hence  $$\langle Ax,y \rangle=(\text{row $i$ of $A$})Ae_j$$
Which is equals to:
$$(\text{row $i$ of $A$})(\text{column $j$ of $A$})$$
Similarly, 
$$\langle x,Ay \rangle =(\text{row $j$ of $A$})(\text{column $i$ of $A$})$$
Here I stopped. 
Is my way correct so far? If so, how can I proceed? 
Thank you very much. 
 A: If $A$ is symmetric then the claim follows readily since
$$\langle\mathbf{x},\mathbf{y}\rangle = \mathbf{x}^\mathrm{T}A\mathbf{y} = \left(\mathbf{x}^\mathrm{T}A\mathbf{y}\right)^\mathrm{T} = \mathbf{y}^\mathrm{T}A^\mathrm{T}\mathbf{x} = \mathbf{y}^\mathrm{T}A\mathbf{x} = \langle\mathbf{y},\mathbf{x}\rangle$$
The second equality follows because $\mathbf{x}^\mathrm{T}A\mathbf{y}$ is $1\times 1$ (a number if you will) which is invariant under transpose. The second last equality follows since we assumed $A$ is symmetric by hypothesis.
Conversely, suppose that the inner product is symmetric. Then
$$\langle\mathbf{e}_i,\ \mathbf{e}_j\rangle = \langle\mathbf{e}_j,\ \mathbf{e}_i\rangle$$
for all $1\le i,\ j\le n$ where we denote $\mathbf{e}_i$ as the $i$th standard basis vector. But note that $\mathbf{e}_i^\mathrm{T}A\mathbf{e}_j$ has the effect of selecting the $ij$th entry of $A$. Therefore
$$(A)_{ij} = \langle\mathbf{e}_i,\ \mathbf{e}_j\rangle = \langle\mathbf{e}_j,\ \mathbf{e}_i\rangle = (A)_{ji}$$
which implies that $A$ is symmetric.
