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i wanted to ask for a clarification.

I was looking around in my linear algebra text when i reached this justification:

Considered two coloumn vectors $X$ and $Y$, and assume $$X ^tC Y = X^t C^t Y$$ for every $X,Y \in V$, with $V$ an $n$-dimensional vectorial space, with $X^t, Y^t$ being the transposed of $ X, Y$.

My book says that because of this is valid for every $X,Y$, we can deduce: $$ C^t = C$$

Now, it is intuitively true, but i was wondering if ,maybe the general sum (it's a bilinear form) could equals without needing $C^t = C$. I've seen this type of justification also in other theorems, but i want to know if there is a way to prove it formally, beacuse i'm not satisfied.

Thanks in advance.

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    $\begingroup$ See what happens if $X=e_i$ and $Y=e_j$ ($i$-th and $j$-th standard basis vectors of $V$). $\endgroup$ Oct 23 at 0:32
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    $\begingroup$ What does this have to do with the “null” in the title? $\endgroup$ Oct 23 at 0:40
  • $\begingroup$ The transpose of a linear transformation doesn’t make sense over a general vector space. But it does make sense if the vector space is given an inner product. $\endgroup$ Oct 23 at 0:43
  • $\begingroup$ Sorry, language translations $\endgroup$ Oct 23 at 0:45
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since these are column vectors, evidently $V=\mathbb F^n$

select $n$ linearly independent vectors $\mathbf x_k$ and $n$ linearly independent $\mathbf y_k$ (if you like you can set $\mathbf y_k:=\mathbf x_k$)

Then for $k,j \in \big\{1,2,...,n\big\}$
$\mathbf x_k^T C\mathbf y_j=\mathbf x_k^T C^T\mathbf y_j$
$\implies\mathbf x_k^T\big( C-C^T\big)\mathbf y_j = \mathbf 0$
$\implies \mathbf X^T\big( C-C^T\big)\mathbf Y = \mathbf 0$
where $\mathbf X$ and $\mathbf Y$ have column $j$ given by $\mathbf x_j$ and $\mathbf y_j$ respectively. But $\mathbf X$ and $\mathbf Y$ are invertible,
$\implies \big( C-C^T\big) = \mathbf 0$
$\implies C=C^T$

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Setting $A = C-C^t$ you have $X^tAY=0$ for all $X,Y$. Let $e_j$ denote the $j$-th standard basis vector. The $(i,j)$-th entry of $A$ can be expressed as $e_i^tAe_j$. But this is zero, hence $A=0$, i.e., $C^t = C$.

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