Prove that set is a basis for a column space

Question

let $A$ be an $m\times n$ matrix and let {$\vec{a_1},...,\vec{a_k},\vec{a_{k+1}},...,\vec{a_n}$} be a basis for $\Bbb{R^n}$ if {$\vec{a_1},...,\vec{a_k}$} is a basis for Null(A) how would you prove that {$A\vec{a_{k+1}},...,A\vec{a_n}$} is a basis for Col(A)

I'm really stuck on this problem, I would like to show that the dimension of Col(A) is $n-k$ and then prove that {$A\vec{a_{k+1}},...,A\vec{a_n}$} is linearly independent. But I have no idea how to proceed and show any of these

Answer 1

HINT

Part 1) $\vec{Aa_{k+1}},...,\vec{Aa_n}\,$ are linearly independent

To show that let consider

$$c_{k+1}A\vec{a_{k+1}}+...+c_nA\vec{a_n}=0 \iff A(c_{k+1}\vec{a_{k+1}}+...+c_n\vec{a_n})=0\\\iff c_{k+1}\vec{a_{k+1}}+...+c_n\vec{a_n} \in \ker(A)\iff c_{k+1}\vec{a_{k+1}}+...+c_n\vec{a_n}=b_{1}\vec{a_{1}}+...+b_k\vec{a_k}\\\iff -b_{1}\vec{a_{1}}-...-b_k\vec{a_k}+c_{k+1}\vec{a_{k+1}}+...+c_n\vec{a_n}=0$$

but we are given that $\vec{a_{1}},...,\vec{a_n}\,$ are linearly independent then $c_i=b_i=0$.

Part 2) $\vec{Aa_{k+1}},...,\vec{Aa_n}\,$ span

Observe that $\forall \vec v\in Col(A)$ we have $\vec v=A \vec w\,$ with $$\vec w=\sum_{i=1}^n b_i \vec a_i \implies \vec v=A\cdot\sum_{i=1}^n b_i \vec a_i\,=\sum_{i=k+1}^n b_i A\vec a_i\,$$

thus $\vec{Aa_{k+1}},...,\vec{Aa_n}\,$ span $\,Col(A)$.

Answer 2

Highlights:

Suppose $\;x_{k+1},...,x_n\in\Bbb R\;$ are such that

$$0=\sum_{i=k+1}^n x_iAa_i\stackrel{\text{lin. A}}=\sum_{i=k+1}^nA(x_ia_i)\stackrel{\text{lin. A}}=A\left(\sum_{i=k+1}^nx_ia_i\right)\implies$$

$$\sum_{i=k+1}^nx_ia_i\in\text{Null}\,A\implies\sum_{i=k+1}^nx_ia_i=\sum_{j=1}^ky_ja_j\;,\;\;\text{for some scalars}\;\;a_1,...,a_k\in\Bbb R\implies$$

$$\sum_{j=1}^ky_ja_j-\sum_{i=k+1}^nx_ia_i=0\ldots\text{ but the vectors}\;a_1,...a_k,a_{k+1},...,a_n\;\;\text{are lin. indep}\ldots$$

...and now you finish the proof (half a line more)