$A$ is symmetric iff $A^T = A$

Question

I saw this question in my linear algebra textbook:

Show $A$ is symmetric if and only $A^T = A$.

I started by doing the following using inner product:

$$\begin{array}{rl}&A \text{ is symmetric}\\ \Leftrightarrow & g(Ax, y) = g(x, Ay)\\ \Leftrightarrow &(Ax)^Ty = x^T(Ay)\\ \Leftrightarrow &(A^Tx^T)y = x^T(Ay)\\ \Leftrightarrow &A^T(x^Ty) = A(x^Ty)\\ \Leftrightarrow &A^T = A \end{array}$$

I don't know how to do the reverse direction. Can someone help me out?

I have read If the inner product is symmetric iff $A=A^{T}$ and Prove A is symmetric matrix iff $A$ is square and $x^T Ay = (Ax)^T y$, but the arguments for the reverse direction don't have any real validity, hence I am asking here.

Edit:

My textbook's definition of a symmetric matrix:

$$\begin{array}{rl}&A \text{ is symmetric}\\ \Leftrightarrow & g(Ax, y) = g(x, Ay)\\\end{array}$$

Answer 1

Some of your steps are incorrect. For example, when manipulating transpose, you should have

$$(AB)^T = B^T A^T.$$

Thus $(Ax)^T = x^TA^T$. Secondly, it is not true that $x^T (Ay) = A(x^Ty)$. You are not allowed to change the order of multiplication when working on matrices. Indeed, you get

$$x^T A^T y = x^T A y$$

for all $x, y$. In particular, you can plug in $x = e_i$ (that is, the vector with $1$ at the $i$-th component and zero otherwise) and $y= e_j$ to conlude that

$$(A^T)_{ij} = A_{ij},$$

thus $A^T$ and $A$ has the same $(i, j)$ entries for all $i, j$. Thus $A^T = A$.

Answer 2

As others have explained, there's a few mistakes in your proposed solution solution. The way I'd do it:

Let $A$ denote a fixed but arbitrary $n \times n$ matrix and bind $x$ and $y$ to $\mathbb{R}^n$. Then:

$$\begin{array}{rl}&A \text{ is symmetric}\\ \iff &\forall x,y[Ax \bullet y = x \bullet Ay]\\ \iff &\forall x,y[(Ax)^Ty = x^T(Ay)]\\ \iff &\forall x,y[x^TA^Ty = x^TAy]\\ \iff &A^T = A \end{array}$$

The only tricky bit is the last equivalence. The backward direction is easy. For the forward direction, use:

Proposition. Consider fixed but arbitrary $m \times n$ matrices $A$ and $B$. Bind $x$ to $\mathbb{R}^m$ and $y$ to $\mathbb{R}^n$. Then:

The statement $\forall x,y[x^T A y = x^T B y]$ implies that $A=B$.

I leave the proof as an exercise, but hint: think about the expression $e_i^T A e_j.$