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Help in order to prove the following , for a given matrix $A \in M_{m \times n}(K)$ where $K$ is a field, I already prove that I have a bilinear form $g_{A}:K^{m} \times K^{n} \to K$ such for $(X,Y) \in K^{m} \times K^{n}$ then $g_{A}(X,Y)=X^{T} A Y$ (in this exercise Im thinking the elements of $K^{m}$ and $K^{n}$ as columns) . So I want to prove that $A$ is a symetric square matrix in $M_{n \times n}(K)$ if and only if $g_{A}$ is a symetric bilinear form.

$\Leftarrow$ If $A$ is a symetric square matrix then I want to prove that for $X,Y \in K^{n}$ then $X^{T}AY=Y^{T}AX$ since $A=A^{T}$ then I got $X^{T}AY=X^{T}A^{T}Y$ and $Y^{T}AX=Y^{T}A^{T}X$ so $(X^{T}AY)^{T}=Y^{T}A^{T}X=Y^{T}AX$ also $(Y^{T}AX)^{T}=X^{T}AY$ so how I can end proving this.

Im stuck in a similar way in the other implication, so any help in order to end proving this equivalence will be apreciated, Thank you.

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    I don't understand where you're stuct. The first implication is done. You proved, that $X^TAY=Y^TAX$ and that was your goal. – 2017-02-14
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    I dont understand why what I did implies $X^{T}AY=Y^{T}AX$ i mean does it happen that $X^{T}AY=(X^{T}AY)^{T}$? if yes, why? I dont see it :(. , also, why $X^{T}AY=Y^{T}A^{T}$? :/ @Tom83B – 2017-02-14

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I think the second implication should be easy, when you realise, what you already have:

  • $X^TAY=Y^TAX$
  • also $X^T AY=Y^TA^TX$

try work it out from here.