Suppose $M$ is a $m\times n$ matrix $(m\geq n)$. Now for any $k$, where $1\leq k\leq n$, in the first $k$ columns the top left $k\times k$ minor has rank $k$ (denote it by $M_k$). And for all $(k+1)\times (k+1)$ minor containing $M_k$ in the first $(k+1)$ columns have determinant 0. Then show that the first $(k+1)$ columns of $M$ are linearly dependent.
Let $M=(a_{i,j})$. Since for all $(k+1)\times (k+1)$ minor containing $M_k$ in the first $(k+1)$ columns have determinant 0 we have for all $k+1\leq i\leq m$ we have $$b_{1,i}\begin{bmatrix}a_{1,1}\\ a_{2,1}\\ \vdots\\ a_{k,1}\\ a_{i,1}\end{bmatrix}+b_{2,i} \begin{bmatrix}a_{1,2}\\ a_{2,2}\\ \vdots\\ a_{k,2}\\ a_{i,2}\end{bmatrix}+\cdots+b_{k,i} \begin{bmatrix}a_{1,k}\\ a_{2,k}\\ \vdots\\ a_{k,k}\\ a_{i,k}\end{bmatrix}+b_{k+1,i} \begin{bmatrix}a_{1,k+1}\\ a_{2,k+1}\\ \vdots\\ a_{k,k+1}\\ a_{i,k+1}\end{bmatrix}=\begin{bmatrix}0\\ 0\\ \vdots\\ 0\\ 0\end{bmatrix}$$ But with this we get a set of $k+1$ coefficients. How can I get coefficients for first $(k+1)$ many $(k+1)\times 1$ columns to show they are linearly dependent
Let $\hat A$ be the $m\times (k+1)$ matrix , which contains the first $k+1$ columns of $A$, We need to show that its rank is equal to $k$. We already know that its rank is at least $k$, since it possesses $k$ linearly independent rows, namely the first $k$ rows. In particular, is $a_j$ are the rows of $\hat A$, then $$ \dim \mathrm{span}(a_1,\ldots,a_k)=k $$
If its rank was equal to $k+1$, then it's row space $W$, i.e. $$ W={\mathrm{span}}(a_1,\ldots,a_{m}) $$ the space spanned by its rows, would have dimension $k+1$. This means that at least one of the $a_{k+2},\ldots,a_m$, say $a_\ell$ does not belong to $\mathrm{span}(a_1,\ldots,a_k)$, and hence $$ a_1,\ldots,a_k,a_\ell, $$ are linearly independent, and hence the determinant of the corresponding $(k+1)(k+1)$ matrix does not vanish.