 Solution

Put $b(v,w)=\phi (w)(v)$. Each $\phi (w)$ belongs to the dual space $V^{\prime }$, so it is a linear functional. For real numbers $a,b_0$ and vectors $v_1,v_2,w$, this givesThe linearity of $\phi$ gives linearity in the second argument as well:Thus $b$ is a bilinear form.

Its right radical of a bilinear form is exactly $\ker \phi$, becauseIf $b$ is a nondegenerate bilinear form, this kernel is zero. Since $V$ is a finite-dimensional vector space and $\dim V^{\prime }=\dim V$, the rank-nullity theorem makes $\phi$ surjective.

Conversely, if $\phi$ is surjective, the same dimension equality makes it injective, so the right radical is zero. If $v$ lies in the left radical, then $f(v)=0$ for every $f\in V^{\prime }$, because every $f$ is some $\phi (w)$. A nonzero $v$ can be included in a basis and detected by its coordinate functional, so this forces $v=0$. Both radicals vanish, proving