# If $A$ is symmetric, why is $x^TA^Ty=(x^TA^Ty)^T$ obvious?

The question has emerged when I read the proof of the following Theorem:

Let $$V$$ be a $$\mathbb{R}$$-vector-space, with $$\dim(V)=n< \infty, C=(c_{1},...,c_{n})$$ a Basis of $$V$$, and $$B$$ a bilinear form. Let $$A$$ be the Gramian matrix with $$a_{ij}=B(c_{i},c_{j})$$ and $$1\leq i,j\leq n$$ . Then the following is true: $$B$$ is symmetric $$\leftrightarrow A$$ is symmetric.

I'm only interessted in the $$\leftarrow$$Direction.

This is the part of the proof that I don't understand completely: $$x,y\in \mathbb{R}^n$$. Assume that $$A$$ is symmetric, then $$A=A^T$$. Then $$x^TAy=x^TA^Ty\overset{?}{=}(x^TA^Ty)^T=y^TAx$$=... the rest of the proof is not important for my question.

I tried to proof the equality $$x^TA^Ty=(x^TA^Ty)^T$$:

$$d:= x^TA^Ty \quad d= x^TA^Ty=x^TAy=(\sum \limits_{i=1}^{n}x_{i}^T a_{i1},...,\sum \limits_{i=1}^{n}x_{i}^T a_{in})\cdot y \\ =\sum \limits_{j=1}^{n}y_{j}\sum \limits_{i=1}^{n}x_{i}^T a_{ij}\overset{y_{j}=y_{j}^T,...}{=} \sum \limits_{j=1}^{n}y_{j}^T\sum \limits_{i=1}^{n}a_{ji}^T x_{i} =\sum \limits_{j=1}^{n}\sum \limits_{i=1}^{n}y_{j}^T a_{ji} x_{i} = \sum \limits_{i=1}^{n}\sum \limits_{j=1}^{n}y_{j}^T a_{ji} x_{i}= \sum \limits_{i=1}^{n}x_{i}\sum \limits_{j=1}^{n}y_{j}^T a_{ji} =(\sum \limits_{j=1}^{n}y_{j}^T a_{j1},...,\sum \limits_{j=1}^{n}y_{j}^T a_{jn})\cdot x = y^TA^Tx=y^TAx=(x^TA^Ty)^T$$

But since they just wrote $$x^TA^Ty=(x^TA^Ty)^T$$in the proof like it would be obvious, I guess there must be a much shorter explanation that makes my proof unnecessary, do you have an idea of such an explanation?

PS: I'm not used to write about math in english, please ask if something doesn't makes sense to you.

• What is the transpose of a $1 \times 1$ matrix? – Rodrigo de Azevedo Mar 8 at 15:56
• $x^TA^Ty$ is just an element of $\Bbb R$... – TonyK Mar 8 at 15:56
• Any 1x1 matrix is it's own transpose. – mathreadler Mar 8 at 16:01

Just note that $$x^T A y$$ is a scalar, or a $$1 \times 1$$ matrix if you will.