# How to show that $T$ is linearly independent?

This question came in my exam, and is by no mean a homework.

Let $\{v_1,v_2\}$ be a linearly independent subset of a vector space $V$ and let $w$ belong to $V$ but not to $\operatorname{Span}(\{v_1,v_2\})$. We need to show that the set $T= \{v_1 + w, v_2 + w\}$ is linearly independent.

My reasoning so far:

$\{v_1,v_2\}$ are linearly independent, thus there exist $c_1$ and $c_2$ such that $c_1v_1 + c_2v_2 = 0$, $c_1=c_2=0$.

$w$ is not in $\operatorname{Span}(\{v_1,v_2\})$, thus the set $\{v_1,v_2,w\}$ is linearly independent.

Thus there exist another $c_1,c_2,c_3$ such that $c_1v_1+c_2v_2+c_3w = 0$.

$c_1v_1+c_2v_2+c_3w = 0 \implies$

$c_3w = -c_1v_1 - c_2v_2 \implies w = (-c_1/c_3)v_1 + (-c_2/c_3)v_3$

$T= \{v_1 + w, v_2 + w\}$. Let there be $d_1$ and $d_2$ scalar such that

$d_1(v_1 + w) + d_2(v_2 + w)= 0$

$\implies d_1(v_1 + (-c_1/c_3)v_1) + d_2(v_2 + (-c_2/c_3)v_3) = 0$

$\implies v_1(d_1 + (-c_1/c_3)d_1) + v_2 (d_2 + (-c_2/c_3)d_2) = 0$,

but $v_1$ and $v_2$ are linearly independent.

Thus, $d_1 + (-c_1/c_3)d_1 = d_2 + (-c_2/c_3)d_2 = 0$

We conclude that the set $T$ is linearly independent.

Is this correct?

EDIT:

$c_1v_1+c_2v_2+c_3w = 0$

T linearly independent, there exit $d_1(v_1 + w) + d_2(v_2 + w)= 0$

$\implies d_1.v_1 + d_1.w + d_2.v_2 + d_2.w = 0$

$\implies (d_1.v_1 + d_2.v_2) + (d_1.w + d_2.w )= 0$

But $d_1.v_1 + d_2.v_2 = 0$ since $v_1$ and $v_2$ are linearly independent

thus $d_1 = d_2 = 0$

$\implies d_1(v_1 + w) + d_2(v_2 + w)= 0$ with $d_1 = d_2 = 0$

T is linearly independent

Note that $v_1,v_2,\ldots,v_k$ are linear independant does not mean (as you write) that there exist $c_1,c_2,\ldots,c_k$ such that $c_1v_1+\ldots +c_kv_k=0$ and $c_1=\ldots =c_k=0$. Rather it menas that for all $c_1,c_2,\ldots,c_k$ with $c_1v_1+\ldots +c_kv_k=0$ we have that $c_1=\ldots =c_k=0$. However, in your reasoning you seem to make use of the correct form.

Nevertheless, you divide by $c_3$ without knowing whether $c_3\ne 0$. Can you modify your calculation to avoid this division?

Apart from this a remark about notation: I am always unhappy when people talk about a set of vectors being linearly independant. For example (not related to the problem statement at hand), assume $\{v_1,v_2\}\subset V$ is linearly indpendant and let $w=v_1+2v_2$. Can it happen that the set $\{v_1,v_2,w\}$ is linearly independent? Maybe surprisingly the answer is: Yes, it can! (Can you find an example for this?)

• I edited my reasoning, is this correct? – abedzantout May 10 '15 at 12:22
• If we take the vectors (1, 1) and (1, 2), then w = (1,1) + 2(1,3) = (3,2). The three vectors are linearly independent. And yes, these kind of question are annoying, and created some problems to my colleagues during their exam. – abedzantout May 10 '15 at 12:24

you have shown that $\{v_1, v_2, w\}$ is linearly independent. you want to show $\{v_1+w, v_2 + w\}$ is linearly independent.

suppose $$c_1(v_1+w) + c_2(v_2 + w) = 0.\tag 1$$ we need to show that $$c_1 = 0, c_2 = 0$$

rewrite $(1)$ as $$c_1v_1 + c_2v_2 + (c_1+c_2)w = 0.$$ use the fact that $\{v_1, v_2, w\}$ is linearly independent to conclude $c_1 = 0, c_2 = 0.$ that concludes the proof.