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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.

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    $\begingroup$ What is the transpose of a $1 \times 1$ matrix? $\endgroup$ – Rodrigo de Azevedo Mar 8 at 15:56
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    $\begingroup$ $x^TA^Ty$ is just an element of $\Bbb R$... $\endgroup$ – TonyK Mar 8 at 15:56
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    $\begingroup$ Any 1x1 matrix is it's own transpose. $\endgroup$ – mathreadler Mar 8 at 16:01
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Just note that $x^T A y$ is a scalar, or a $1 \times 1$ matrix if you will.

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    $\begingroup$ I actually used that in my proof :D Looks like I wasn't able to see the forest for the trees ( I hope the translation of the phrase is correct). $\endgroup$ – CherryBlossom1878 Mar 8 at 16:03

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