Let $\phi$ a bilinear form. Then $\phi(Ax,y)=\phi(x,A^Ty)$ for each $A \in M_n(\mathbb{K})$
I assumed that this statement is true (I couldn't find counterexamples). So I tried proving this by the properties of bilinear forms:
$\phi(u+v,w)= \phi(u,w) + \phi(v,w)$ and $\phi(\lambda u, v) = \lambda \phi(u,v)$
$\phi(u,v+w)= \phi(u,v) + \phi(u,w)$ and $\phi(u, \lambda v) = \lambda \phi(u,v)$
I tried to transform this equations to prove my task. I'm stuck here.
Can you give me a hint?
Let $\varphi:\mathbb{K}^n\times\mathbb{K}^n\rightarrow\mathbb{K}$ a bilinear form and let $M:=(\varphi(e_i,e_j))_{1\leqslant i,j\leqslant n}$ where $(e_i)_{1\leqslant i\leqslant n}$ is a base of $\mathbb{K}^n$. Thus for $(X,Y)\in\mathbb{K}^n\times\mathbb{K}^n$, if $X=\sum_{i=1}^{n^2}X_i e_i$ and $Y=\sum_{i=1}^{n^2}Y_ie_i$ we have $$ \varphi(X,Y)=\sum_{1\leqslant i,j\leqslant n^2}X_iY_j\varphi(e_i,e_j)=X^TMY $$ It follows that for all $(X,Y)\in\mathbb{K}^n\times\mathbb{K}^n$ and $A\in\mathcal{M}_n(\mathbb{K})$ we have $$ \varphi(AX,Y)=(AX)^TMY=X^TA^TMY $$ and $$ \varphi(X,A^TY)=X^TMA^TY $$ thus $\varphi(AX,Y)=\varphi(X,A^TY)$ only if $A^TM=MA^T$.