Proving vectors as a basis in $E^{m}$ Show that if the vectors $a_{1}$, $a_2$, $\cdots$, $a_m$, are a basis in $E^{m}$, the vectors $a_{1}$, $a_2$, $\cdots$, $a_{p-1}$, $a_{q}, a_{p+1}, \cdots,a_{m}$, also are a basis if and only if $y_{p,q} \neq 0$, where $y_{p,q}$ is defined by the following tableau:
\begin{matrix}
1& 0& \cdots & 0& y_{1,m+1}   & y_{1,m+2}& \cdots & y_{1n} & y_{10}\\
0& 1& \cdot  & 0&y_{2,m+1}& y_{2,m+2}& \cdots & y_{2n} & y_{20}\\
0& 0& \cdot  & 0& \cdot    & \cdot    & \cdot  & \cdot  & \cdot\\
\vdots& \vdots& \vdots  & \vdots& \vdots& \vdots& \vdots & \vdots& \vdots\\
0& 0& \cdot  & 1& y_{m,m+1} & y_{m,m+2} & \cdots & y_{mn} & y_{m0}
 \end{matrix}
Can the necessary and sufficient conditions be defined as follows.
If $a_{1}$, $a_2$, $\cdots$, $a_{p-1}$, $a_{q}, a_{p+1}, \cdots,a_{m}$ are a basis in $E^{m}$ then $y_{p,q} \neq 0$ which implies to prove that they're LI (necessary condition) and if $y_{p,q} \neq 0$ then  $a_{1}$, $a_2$, $\cdots$, $a_{p-1}$, $a_{q}, a_{p+1}, \cdots,a_{m}$ are a basis in $E^{m}$ (sufficient condition)?
Does anyone have any idea to prove this? Any hint is welcome.
Thanks.
 A: Remark: Your problem has nothing to be with the rightmost column
$(y_{1,0},y_{2,0},\dots,y_{m,0})^T$.  I'll omit that column from the matrix.
Settings
Let $A = \begin{bmatrix}\mathbf{a}_1&\mathbf{a}_2&\cdots&\mathbf{a}_m&|\mathbf{a}_{m+1} & \cdots&\mathbf{a}_n\end{bmatrix}$ be the original coefficient matrix in $E^{m \times n}$, where $\mathbf{a}_j$ denotes the $j$-th column of $A$.  By changing $A$ to the given matrix $\begin{bmatrix}I_m&|\mathbf{y}_{m+1} & \cdots&\mathbf{y}_n\end{bmatrix}$, where $\mathbf{y}_j$ denotes the $j$-th column of the given matrix (i.e. $\mathbf{y}_j = \mathbf{e}_j \;\forall j \in \{1,\dots,n\}$ and $\mathbf{y}_j^T = (y_{1,j},y_{2,j},\dots,y_{m,j}) \;\forall j \in \{n+1,\dots,m\}$), using row operations, the given matrix equals $B^{-1}A$, where $B := \begin{bmatrix}\mathbf{a}_1&\mathbf{a}_2&\cdots&\mathbf{a}_m\end{bmatrix}$ is formed by the given basis.  To see this, use the fact that
\begin{align}
B^{-1} B &= I \\
B^{-1} \begin{bmatrix}\mathbf{a}_1&\mathbf{a}_2&\cdots&\mathbf{a}_m\end{bmatrix} &= \begin{bmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \cdots & \mathbf{e}_m \end{bmatrix}
\end{align}
Therefore, we have $B^{-1}\mathbf{a}_j = \mathbf{y}_j \;\forall j \in \{1,2,\dots,n\}$.  This is our starting point.
Sufficient condition
Assume that $y_{pq} \ne 0$.
\begin{align}
B^{-1}\mathbf{a}_q &= \mathbf{y}_q \\
\mathbf{a}_q &= B \mathbf{y}_q \\
&= \sum_{i = 1}^m y_{i,q} \mathbf{a}_i \\
&= y_{p,q} \mathbf{a}_p + \sum_{\substack{i = 1 \\ i \ne p}}^m y_{i,q} \mathbf{a}_i \\
\frac{1}{y_{p,q}} \mathbf{a}_q &= \mathbf{a}_p + \sum_{\substack{i = 1 \\ i \ne p}}^m \frac{y_{i,q}}{y_{p,q}} \mathbf{a}_i \\
\mathbf{a}_p &= \frac{1}{y_{p,q}} \mathbf{a}_q - \sum_{\substack{i = 1 \\ i \ne p}}^m \frac{y_{i,q}}{y_{p,q}} \mathbf{a}_i
\end{align}
Since it's given that $\left\{ \mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_m \right\}$ is a basis for $E^m$, then $$\left\{ \mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_{p - 1}, \frac{1}{y_{p,q}} \mathbf{a}_q - \sum_{\substack{i = 1 \\ i \ne p}}^m \frac{y_{i,q}}{y_{p,q}} \mathbf{a}_i, \mathbf{a}_{p + 1}, \dots, \mathbf{a}_m \right\},$$ is a basis for $E^m$.  Hence $\left\{ \mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_{p - 1}, \mathbf{a}_q, \mathbf{a}_{p + 1}, \dots, \mathbf{a}_m \right\}$ is a basis for $E^m$.
Necessary condition
Suppose that $y_{pq} = 0$.  Then from the section above, we have
$$\mathbf{a}_q = y_{p,q} \mathbf{a}_p + \sum_{\substack{i = 1 \\ i \ne p}}^m y_{i,q} \mathbf{a}_i = \sum_{\substack{i = 1 \\ i \ne p}}^m y_{i,q} \mathbf{a}_i,$$ so $\mathbf{a}_q$ can be represented by $\left\{ \mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_{p - 1}, \mathbf{a}_{p + 1}, \dots, \mathbf{a}_m \right\}$.  In other words, the vector $\mathbf{a}_q$ has two different representations by the basis $\left\{ \mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_{p - 1}, \mathbf{a}_q, \mathbf{a}_{p + 1}, \dots, \mathbf{a}_m \right\}$.  This is a contradiction.
