I have seen a few answers to this question in this site, one of them deals with a very special case, while the other one only one side of the problem is proved, so I wanted a better answer and asked this question.
Let $f$ be a bilinear form on a finite dimensional vector space $V$ over some field $\mathbb F$. I was trying to prove that $f$ can be expressed as a product of two linear functionals i.e. $$f(u,v) = L_{1}(u) L_{2}(v) \text{for} L_{1},L_{2} \in V^*$$ if and only if $f$ has rank $1$.
I just know the definitions. If someone suggests an answer it will be very much helpful.
Note that the end of the statement should be “has rank $1$ or $0$”.
Suppose that there are linear functionals $L_1,L_2\in V^*$ such that$$\tag{1}\bigl(\forall (u,v)\in V\bigr):f(u,v)=L_1(u)L_2(v).$$Fix a basis $B=\{e_1,e_2,\ldots,e_n\}$ of $V$. Then there are elements $a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n$ of $\mathbb F$ such that, whenever $\alpha_1,\alpha_2,\ldots,\alpha_n\in\mathbb F$,$$L_1(\alpha_1e_1+\alpha_2e_2+\cdots+\alpha_ne_n)=\alpha_1a_1+\alpha_2a_2+\cdots+\alpha_na_n$$and$$L_2(\alpha_1e_1+\alpha_2e_2+\cdots+\alpha_ne_n)=\alpha_1b_1+\alpha_2b_2+\cdots+\alpha_nb_n.$$Therefore, the matrix of $f$ with respect to $B$ is$$\begin{bmatrix}a_1b_1&a_1b_2&\cdots&a_1b_n\\a_2b_1&a_2b_2&\cdots&a_2b_n\\\vdots&\vdots&\ddots&\vdots\\a_nb_1&a_nb_2&\cdots&a_nb_n\end{bmatrix}=\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix}.\begin{bmatrix}b_1&b_2&\cdots&b_n\end{bmatrix},$$which has rank $1$ or $0$.
On the other hand, if the matrix of $f$ with respect to $B$ has rank $1$ or $0$, then the columns are all multiples of a single column. That is, there are elements $\lambda_1,\lambda_2,\ldots,\lambda_n,c_1,c_2,\ldots,c_n$ of $\mathbb F$ such that the matrix of $f$ with respect to $B$ is$$\begin{bmatrix}\lambda_1c_1&\lambda_1c_2&\cdots&\lambda_1c_n\\\lambda_2c_1&\lambda_2c_2&\cdots&\lambda_2c_n\\\vdots&\vdots&\ddots&\vdots\\\lambda_nc_1&\lambda_nc_2&\cdots&\lambda_nc_n\end{bmatrix}$$and then, if you define$$L_1(\alpha_1e_1+\alpha_2e_2+\cdots+\alpha_ne_n)=\alpha_1\lambda_1+\alpha_2\lambda_2+\cdots+\alpha_n\lambda_n$$and$$L_2(\alpha_1e_1+\alpha_2e_2+\cdots+\alpha_ne_n)=\alpha_1c_1+\alpha_2c_2+\cdots+\alpha_nc_n,$$you will have $(1)$.