Question

Prove that the following are equivalent:

a) The set U = {u^1, u^2,..., u^n} is linearly independent;
b) u^1 does not equal 0 and no u^n is a linear combination of the members of U that precede it in the list;
c) no u^n is a linear combination of the other members of U.

Answer 1

1 => 2

$\{u_{1},\dots,u_{n}\}$ - linearly independent. Let we have $\exists u_{i + 1} = au_1 + bu_2 + \ldots +cu_i\neq 0$

Then $au_{1} + bu_{2} + \ldots + cu_{i} + (-1)u_{i + 1} = 0$ , so there is nonzero scalar combination that subsystem of U is linearly dependent. Then U itself is linearly dependent. - contradiction. So, 2) holds.

2) => 3)

Let we have $\exists u_{i} = au_{1} + bu_{2} + \ldots +cu_{i - 1} + du_{i + 1} + \ldots +eu_{n}\neq 0$

Then if $e \neq 0$ then $\exists u_{n} = (-e^{-1})(au_{1} + bu_{2} + \ldots + cu_{i-1} - u_{i} + du_{i+1} + \ldots + fu_{n-1}) \neq 0$ - so $u_{n}$ is linear

combination of elements $\{u_{1},\dots,u_{n - 1}\}$ . - contradiction to 2).

Thus similar observation leads us to fact that all coefficients near $u_{i+1}, \ldots, u_n$ are zeros, and thus

$u_{i} = au_{1} + bu_{2} + \ldots +cu_{i - 1}\neq 0$

Last fact contradicts 2) again, and so 3) holds.

3) => 1)

If system $\{u_1,\dots,u_n\}$ is linearly dependent then one of the vectors is a linear combination of other ones, but this is impossible since 3) holds, so U is linearly independent.