# Rank $1$ bilinear form is a product of two linear functionals on a finite dimensional vector space.

Let $\mathbb{f}$ be a non-zero bilinear form on a finite dimensional vector sppace $V.$ Then have to show that $\mathbb{f}$ can be expressed as a product of two linear functionals i.e., $\mathbb{f}(\alpha, \beta)=L_1(\alpha)L_2(\beta)$ for $L_i \in V^*$ iff $\mathbb{f}$ has rank $1.$

I proved that if $\mathbb{f}$ is product of two linear functional then its rank is $1$ using left operator and the right operator. I am looking for the proof of the other direction. Help me. Many thanks.

• How did you prove that direction, can you explain a bit?
– user422112
Oct 23, 2017 at 14:31
• It follows from the two induced linear maps from $V$ to $V^{*}$. You can prove that both these linear maps has same nullity, and hence the same rank, which we define as the rank of a bilinear form. Mar 18, 2019 at 7:25

Let $n={\mathsf{dim}}(V)$. Consider the left and right versions of $f$, $f_L : V \to V^{*}$ defined by $f_L(x)(y)=f(x,y)$ and $f_R : V \to V^{*}$ defined by $f_R(x)(y)=f(y,x)$. We know that $f_L$ and $f_R$ both have rank $1$. So ${\mathsf Ker}(f_L)$ has a basis made of $n-1$ vectors, say $u_2,u_3,\ldots,u_n$. Similarly ${\mathsf Ker}(f_R)$ has a basis made of $n-1$ vectors, say $v_2,v_3,\ldots,v_n$. Next, let $u_1\not\in{\mathsf Ker}(f_L)$ and $v_1\not\in{\mathsf Ker}(f_R)$, and $c=f(u_1,v_1)$. Then $c\neq 0$ (otherwise $f$ is zero everywhere) $(u_1,u_2,\ldots,u_n)$ and $(v_1,v_2,\ldots,v_n)$ are bases of $V$. Define $L_1\in V^*$ by $L_1(u_1)=1$ and $L_1(u_k)=0$ for $k>1$. Define also $L_2\in V^*$ by $L_2(v_1)=c$ and $L_2(v_k)=0$ for $k>1$. Then $f(u_i,v_j)=L_1(u_i)L_2(v_j)$ for any $i,j$, and the result follows by bilinearity.
• Could you explain why $f_{L}, f_{R}$ are of rank $1$? I'm not seeing it.
• @AJY $f_L,f_R$ are of rank $1$ because $f$ is Jun 12, 2018 at 4:43