Let \(f\) be a nonzero bilinear form on a finite-dimensional vector space \(V\). Show that \(f\) can be expressed as a product of two linear functionals (i.e., \(f(\alpha, \beta) = L_1(\alpha)L_2(\beta)\) for \(L_1, L_2\) in \(V^*\)) if and only if \(f\) has rank 1.
The printed exercise does not say that \(f\) is nonzero. Without that qualification the assertion is false: the zero form is \(0(\alpha)L_2(\beta)\) for any \(L_2\), but has rank zero. Equivalently one could leave \(f\) arbitrary and require both factor functionals to be nonzero.
Solution
Define \[ L_f:V\longrightarrow V^*, \qquad L_f(\alpha)(\beta)=f(\alpha,\beta). \] Then \[ \operatorname{rank}f=\dim\operatorname{im}L_f. \]
1. A product of two functionals has rank one when the form is nonzero.
Suppose \[ f(\alpha,\beta)=L_1(\alpha)L_2(\beta). \] Because \(f\ne0\), neither \(L_1\) nor \(L_2\) is zero. For every \(\alpha\), \[ L_f(\alpha)=L_1(\alpha)L_2, \] so \[ \operatorname{im}L_f\subseteq\operatorname{span}\{L_2\}. \] Since \(L_1\ne0\), choose \(\alpha_0\) with \(L_1(\alpha_0)\ne0\). Then \[ L_f(\alpha_0)=L_1(\alpha_0)L_2\ne0, \] so the image is not zero. Hence \[ \operatorname{im}L_f=\operatorname{span}\{L_2\} \] is one-dimensional, and \[ \operatorname{rank}f=1. \]
2. Every rank-one bilinear form factors as a product of two functionals.
Assume \[ \operatorname{rank}f=1. \] Then \(\operatorname{im}L_f\) is a one-dimensional subspace of \(V^*\). Choose a nonzero functional \(L_2\) spanning it. For each \(\alpha\in V\) there is a unique scalar \(L_1(\alpha)\) such that \[ L_f(\alpha)=L_1(\alpha)L_2. \]
We claim that \(L_1\) is linear. If \(a,b\) are scalars and \(\alpha,\gamma\in V\), linearity of \(L_f\) gives \[ \begin{aligned} L_1(a\alpha+b\gamma)L_2 &=L_f(a\alpha+b\gamma)\\ &=aL_f(\alpha)+bL_f(\gamma)\\ &=(aL_1(\alpha)+bL_1(\gamma))L_2. \end{aligned} \] Since \(L_2\ne0\), equality of these scalar multiples implies \[ L_1(a\alpha+b\gamma) =aL_1(\alpha)+bL_1(\gamma). \] Thus \(L_1\in V^*\).
Finally, evaluating the identity \(L_f(\alpha)=L_1(\alpha)L_2\) at \(\beta\) gives \[ \boxed{ f(\alpha,\beta)=L_1(\alpha)L_2(\beta)} \] for all \(\alpha,\beta\in V\).