Finding Linearly Independent Vectors in Subspaces

[Faiq]

Homework Statement

The vectors $a_1, a_2, a_3, b_1, b_2, b_3$ are given below
$$\ a_1 = (3~ 2~ 1 ~0) ~~a_2 = (1~ 1~ 0~ 0) ~~ a_3 = (0~ 0~ 1~ 0)~~ b_1 = (3~ 2~ 0~ 2)~~ b_2 = (2 ~2~ 0~ 1)~~ b_3 = (1~ 1~ 0~ 1) $$
The subspace of $\mathbb R^4$ spanned by $a_1, a_2, a_3$ is denoted by $V_1$ and the subspace of $\mathbb R^4$ spanned by $b_1, b_2, b_3$ is denoted by $V_2$

The set of vectors which consists a zero vector and all vectors which belong to $V_1$ and $V_2$ is denoted by $W$.

Write down two linearly independent vectors which belong to $W$

The Attempt at a Solution

Can somebody please explain how to get the independent vectors? I am very aware of the definition of independent vectors. However, I cannot seem to relate the definition and use it to solve this question.

 

[FactChecker]

Hint: a₁, a₂, and a₃ all have 0 as their 4'th coordinate, so they can not generate any vector with a nonzero 4'th coordinate.

 

[Faiq]

And $b_1, b_2, b_3$ all have 0 as their 3rd coordinate, so they cannot generate any vector with a nonzero 3rd coordinate
Thus the resulting vector should be $(p~q~0~0)$
Correct?

 

[FactChecker]

And $b_1, b_2, b_3$ all have 0 as their 3rd coordinate, so they cannot generate any vector with a nonzero 3rd coordinate
Thus the resulting vector should be $(p~q~0~0)$
Correct?

You need to find two vectors. Look for one in V₂ that could not be a linear combination of the a_is and look for a vector in V₁ that could not be a linear combination of the b_is

 

[Faiq]

a₃ and b₃?
And the reasoning behind the special criteria for choosing the vector is this right? If I were to choose a vector in V₁ which was a linear combination of V₂ vectors, then I would technically be choosing just V₂ vectors and not taking into account the V₁ vectors.

 

[FactChecker]

a3 and b3?

In homework, you would want to briefly explain why the two you pick are in W and are linearly independent. Use the definition of linearly independent.

 

[Faiq]

Thank you very much for your help