Proof that a set of vectors is linearly independent

Question

Let $u_1,\ldots,u_k \in \mathbb R^n$, where $k\leq n$

Show that:

If $|u_{jj}| > \sum\limits_{i \neq j}|u_{ij}|$ (sum over the i's) then the vectors are linearly independent ( $u_{ij}$ denotes the $i$-th coordinate of the $j$-th vector).

I tried to prove by contradiction but with no success.

Answer 1

Suppose they are linearly dependent.

Then there would exist some combination.

$a_1 u_1 + a_2 u_2 + \cdots a_k u_k = 0$

Choose the $a_j$ with the largest absolute value.

$|a_j u_{jj}| > \sum_\limits{i\ne j} |a_i u_{ji}|$

$a_1 u_1 + a_2 u_2 + \cdots a_k u_k \ne 0$

Answer 2

make a set of $v_1 ...v_n \in R^k$ where $v_ij=u_ji$

if we can show $v_1 ...v_k$ are linearly independent, we can conclude that the rank is k and thus $u_1 ...u_k$ are linearly independent.

For $v_1 ...v_k$, we may be able to use Doug M's method of contradiction.