I wanna show the following argument:
$X,Y$ are finite-dimensional vector spaces (same scalar field), and $\phi$ is a bilinear mapping from $X\times Y$ to the scalar field.
How can I construct linear mappings $\theta_1,...,\theta_n$ and $\psi_1,...,\psi_n$ such that $\phi(x,y)=\Sigma_{i=1}^n \theta_i(x)\psi_i(y)$ by considering bases for $X$ and $Y$?
and my answer is following:
Note we can define linear mappings $L_\phi:X \rightarrow Y^\#$($Y^\#$: dual of $Y$) and $R_\phi:Y\rightarrow X^\#$ from $\phi$ by $L_\phi(x)(y)=\phi(x,y)=R_\phi(y)(x)$
Let $\{y_1,...,y_m\}$ be a basis of $Y$ and $F=\{f_1,...,f_m\}$ be the coordinate functional of the basis, then $\phi(x,y)=L_\phi(x)(y)=\Sigma_{i=1}^m L_\phi(x)(y_i)f_i(y)=\Sigma_{i=1}^m R_\phi(y_i)(x)f_i(x)$, so by setting $R_\phi(y_i)=\theta_i, f_i=\psi_i$, we have $\phi(x,y)=\Sigma_{i=1}^m \theta_i(x)\psi_i(y)$
I think here $X$ need not be finite dimensional, so there is something wrong. What's wrong here?
There's nothing wrong — your proof is correct. It's true that the conclusion holds under more general assumptions (i.e. when any of $X$, $Y$ is finite dimensional).