How do these column vectors [1,-3, 0], [-2 9 0], [0, 0, 0], and [0, -3, 5] not form a basis?

Question

I am given a question asking if the set of those 4 questions form a basis. The answer is that they do not.

I ignore the third vector which represents the origin, and because there is a 0 in the first row of the fourth vector, it must be linearly independent from the first 2 vectors.

Regarding the first 2 vectors, how can they be linearly dependent? Those two are not multiples of each other.

Answer 1

$0[1,-3,0]+0[-2,9,0]+1[0,0,0]+0[0,-3,5]=[0,0,0]$ but the coefficients are not all zeros.

Answer 2

To be a basis for a space of dimension n a set of vector need to fullfil the following condition

• the vectors in the set are exactly n
• the vectors are linearly independent

In this case you have 4 vectors with n=3 and thus the set can't be a basis, moreover the vector $(0,0,0)$ is a trivial multiple of every vector thus it cant belong to any basis.

Excluding the zero vector you obtain a basis for $\mathbb{R^3}$.

Answer 3

Your mistake is precisely excluding the zero vector. Without it, you have a basis; with it - you don't.

Recall that the basis must be linearly independent. A zero vector makes it linearly dependent. Generally, if you have vectors $\mathbb{0},v_2,v_3,\ldots,v_n$, those are never linearly independent because $1\cdot \mathbb{0}+0\cdot v_2+0\cdot v_3+\cdots+0\cdot v_n=0$ but $1\ne 0$.