Linear Independence of abstract vectors

Question

I've seen this asked many times in problems and don't understand how to concretely prove it.

True or False? Explain the assertion or find a counterexample. If $\{u_1, u_2, u_3\}$ is a linearly independent set in some vector space $V$, then also the set $\{u_1, u_1 + u_2, u_1 + u_2 + u_3\}$ is linearly independent.

Does anybody know the proper way to prove this?

Thank you!

Answer 1

Hint: proceed according to the definition, and build a linear combination of the three vectors. Hence $$ \alpha u_1 + \beta (u_1+u_2) + \gamma (u_1+u_2+u_3) =0. $$ Then collect, and try to use the indipendence of $\{u_1,u_2,u_3\}$.

Answer 2

We can prove by definition looking for $a,b,c\neq 0$ such that

$$ax_1+b(x_1+x_2)+c(x_1+x_2+x_3)=0$$

$$\iff (a+b+c)x_1+(b+c)x_2+cx_3=0$$

and since $x_1,x_2,x_3$ are linearly independent finding

• $a+b+c=0$
• $b+c=0$
• $c=0$

which is true if and only if $a=b=c=0$ thus also the set $\{u_1, u_1 + u_2, u_1 + u_2 + u_3\}$ is linearly independent.

More directly we can easily see that we can't obtain $u_1 + u_2 + u_3$ from $u_1$ and $u_1 + u_2$ since they don't contain $u_3$ and also $u_1$ and $u_1 + u_2$ are clearly linearly independent.

Answer 3

Hint: Do it indirectly. Assume $\{u_1,u_1+u_2,u_1+u_2+u_3\}$ is linearly dependent and conclude that $\{u_1,u_2,u_3\}$ is linearly dependent.

Answer 4

Take $a_1,a_2,a_3$ such that $a_1u_1+a_2(u_1+u_2)+a_3(u_1+u_2+u_3)=0$. This means that$$\left\{\begin{array}{l}a_1+a_2+a_3=0\\a_2+a_3=0\\a_3=0.\end{array}\right.$$But the only solution of this system is $a_1=a_2=a_3=0$.