If we have $n-1$ dependent vectors of size $n$, why if we remove the $i$th element from each, they're still dependent?

Question

I was reading this answer to the problem of showing $\operatorname{rank}(\operatorname{adj}A)\in\{0,1,n\}$. In the first case we see:

If $A$ has rank less than or equal to $n - 2$ then any $n - 1$ columns of $A$ are linearly dependent and in particular, any $(n-1) \times (n-1)$ submatrix of $A$ has linearly dependent columns.

I can't understand why is that correct. Is there an elementary proof to this?

Is it because for checking that those $n-1$ vectors of size $n$ are linearly dependent, we solve a $n-1$ variables and $n$ equations system and we find a non-zero answer, solving any $n-1$ variables and $n-1$ equations system from the first linear system of equations must also have a non-zero answer?

Answer 1

If you have some relation between the vectors, i.e. some non-trivial linear combination equating zero, the same combination will stay equal to zero when you remove a coordinate.

In other words, suppose that $v_1,v_2,\ldots,v_{n-1}$ are $n-1$ vectors in ${\mathbb R}^n$, and suppose that they are linearly dependent i.e. we have $\lambda_1v_1+\ldots+\lambda_{n-1}v_{n-1}=0$ for some numbers $\lambda_1,\ldots,\lambda_n$ not all zero. If you denote by $p$ the linear projection ${\mathbb R}^n \to {\mathbb R}^{n-1}$ which removes the $i$-th coordinate, then you have

$$ 0=p(0)=p(\lambda_1v_1+\ldots+\lambda_{n-1}v_{n-1})= \lambda_1p(v_1)+\ldots+\lambda_{n-1}p(v_{n-1}) $$

so the vectors $p(v_1),p(v_2),\ldots,p(v_{n-1})$ are linearly dependent also.

Answer 2

Let those $n-1$ vectors be $v_1,...,v_{n-1}$ and $v_k(j)$ denote the $j$-th entry of $v_k$, the condition indicates that there are nontrivial coefficients $c_k$ such that $\sum c_k v_k(j)=0$ simultaneously for all $j$ in $\{1,..,n\}$.

So the equation is also true if we restrict the range of $j$ from $\{1,..,n\}$ into any subset.

Answer 3

Suppose given any $(n-1)\times (n-1)$ submatrix $$ \tilde{A}=(\tilde{v}_{l_1}, \tilde{v}_{l_2},\cdots, \tilde{v}_{l_{n-1}}) $$ its corresponding column is $v_{l_1}, v_{l_2},\cdots, v_{l_{n-1}}$. Since $v_{l_1}, v_{l_2},\cdots, v_{l_{n-1}}$ are linearly dependent, exists $\lambda_{l_1}, \cdots, \lambda_{l_{n-1}}\in \mathbb{F}$ s.t. $$ \lambda_{l_1}v_{l_1}+ \cdots+ \lambda_{l_{n-1}}v_{l_{n-1}}=0.$$ Then $$ \lambda_{l_1}\tilde{v}_{l_1}+ \cdots+ \lambda_{l_{n-1}}\tilde{v}_{l_{n-1}}=0. $$