= Solution
{c}
All $n$ <singular values> are positive, so $\Sigma$ and $V$ are <invertible matrices> and $A$ has <matrix rank> $n$. The <image of a linear map> is therefore
$$
\boxed{\operatorname{range}(A)=\operatorname{span}\{u_1,\ldots,u_n\},}
$$
with those columns furnishing an <orthonormal basis>. Also $Ax=0$ implies $\Sigma V^Tx=0$ and hence $x=0$. Thus
$$
\boxed{\ker A=\{0\},\qquad\text{its basis is the empty set}.}
$$
Finally $A^Tz=V\Sigma U^Tz=0$ exactly when $U^Tz=0$. The remaining columns of the full <orthogonal matrix> span this <orthogonal complement>:
$$
\boxed{\ker A^T=\operatorname{range}(U_\perp),\qquad
\text{a basis consists of the }m-n\text{ columns of }U_\perp.}
$$
If $m=n$, this last <null space> also equals $\{0\}$ and has the empty basis.
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