A question I'm working on asks me to prove or disprove the following statement if $A$ is $m \times n$ matrix such that {$\vec{v_1},...,\vec{v_k}$} is a basis for $Null(A)$ and if the basis is extended such that {${\vec{v_1},...,\vec{v_k},\vec{v_{k+1}},...,\vec{v_n}}$} forms a basis for $\Bbb{R}^n$ then {$\vec{v_{k+1}},...,\vec{v_n}$} is a basis for $Row(A)$
this is my attempt at a proof which im not sure is correct so dim$Null(A)=k \leq n$ then Rank(A)=$n-k$ = dim Row(A) thus since {$\vec{v_{k+1}},...,\vec{v_n}$} is a set of $n-k$ linearly independent vectors it must span $Row(A)$ and thus is a basis.
Would this proof be valid or am I missing something?
A counterexample: Let $A=\operatorname{diag}(1,0)$. Its null space is obviously spanned by $(0,1)^T$. I can extend this to a basis of $\mathbb R^2$ with $(1,1)^T$, but this vector clearly doesn’t lie in $A$’s row space.
You’ve only shown that the vectors used to extend the null space basis form a basis for some subspace of $\mathbb R^n$ with the correct dimension, but not that they span the correct subspace: they must span the orthogonal complement of $A$’s null space, a.k.a. the column space of $A^T$.