= Solution
{c}
Put $b(v,w)=\varphi(w)(v)$. Each $\varphi(w)$ belongs to the <dual space> $V'$, so it is a <linear functional>. For <real numbers> $a,b_0$ and <vectors> $v_1,v_2,w$, this gives
$$
b(av_1+b_0v_2,w)=\varphi(w)(av_1+b_0v_2)
=a\,b(v_1,w)+b_0b(v_2,w).
$$
The <linearity> of $\varphi$ gives <linearity> in the second argument as well:
$$
b(v,aw_1+b_0w_2)
=(a\varphi(w_1)+b_0\varphi(w_2))(v)
=a\,b(v,w_1)+b_0b(v,w_2).
$$
Thus \b[$b$ is a <bilinear form>].
Its <right radical of a bilinear form> is exactly $\ker\varphi$, because
$$
b(v,w)=0\text{ for every }v\in V
\ \Longleftrightarrow\ \varphi(w)=0.
$$
If $b$ is a <nondegenerate bilinear form>, this <kernel> is zero. Since $V$ is a <finite-dimensional vector space> and $\dim V'=\dim V$, the <rank-nullity theorem> makes $\varphi$ <surjective>.
Conversely, if $\varphi$ 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 $f\in V'$, because every $f$ is some $\varphi(w)$. A nonzero $v$ can be included in a <basis> and detected by its <coordinate functional>, so this forces $v=0$. Both radicals vanish, proving
$$
\boxed{b\text{ is nondegenerate}\ \Longleftrightarrow\ \varphi\text{ is surjective}.}
$$
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