Prove $A_1, A_2, \ldots,A_{\ell-1}, A_{\ell+1}\ldots \ldots, A_{m} $ in $\mathbb R^m$ are linearly independent viewed as vectors in $\mathbb R^{m-1}$? Suppose I have linearly independent vectors $A_1, A_2, \ldots, A_m$ in $\mathbb R^m$
Consider the matrix $B = [A_1, A_2, \ldots  A_m]$ consisting of these vectors and suppose $B^{-1}[A_1, A_2, \ldots,A_{\ell-1}, A_{\ell+1}\ldots \ldots, A_{m}]$ has the $\ell$-row consisting only of $0$'s.
Is it then neccesarily true that $A_1, A_2, \ldots,A_{\ell-1}, A_{\ell+1}\ldots \ldots, A_{m}$ are linearly independent in $\mathbb R^{m-1}$ considered as vectors with the $\ell$-component removed ?
I've tried to prove that $A_1, A_2, \ldots,A_{\ell-1}, A_{\ell+1}\ldots \ldots, A_{m}$, with the $\ell$-component removed, span $\mathbb R^{m-1}$ or are linearly independent, however I've not yet come to a final result.
 A: This is not true. Let $m=3$ and $A_1=(0,2,3),~ A_2=(1,1,1),~ A_3=(0,1,1)$. Then
\begin{align}
\det(A_1,A_2,A_3)=\begin{vmatrix}
0 & 1 & 0\\
2 & 1 & 1\\
3 & 1 & 1
\end{vmatrix}
=-1\neq 0.
\end{align}
If we delete the first components of $A_2$ and $A_3$ the system $\{A_2,A_3\}$ wont be linearly independent in $\mathbb R^2$
Let $A_1=(1,2,3),~A_2=(0,1,1),~A_3=(-1,1,1)$. Then 
$$B=\begin{bmatrix}
1&0&-1\\ 2&1&1\\ 3&1&1
\end{bmatrix}\quad \text{and}\quad 
B^{-1}=\begin{bmatrix}
0&-1&1\\ 1&4&-3\\ -1&-1&1
\end{bmatrix}.$$
The first row of the matrix $B^{-1}[A_2,A_3]$ is only 0 and the system $\{A_2,A_3\}$ after deleted the first components is not linearly independent.
A: I can't understand your question. Why you suppose $ B^{-1}[A_1, A_2, \ldots, A_{l-1}, A_{l-1},\ldots,A_m]$has the$ {l}$ row consisting only of 0's?
It is evident that such vectors are linearly independent in $\mathbb R^{m-1}$, or you can just let $a_l=0$ to prove that the original vectors are also linearly dependent. 
A: It is necessary that any subset is linearly independent so that a set is linearly independent (by definition).
