Basis and dimension of the span of the vectors (0, 0, 0), (9, 0, 0), (8, 1, 0), (1, 8, 9)

Question

Find a basis for the given subspace by deleting linearly dependent vectors. $S = \text{span}\{(0, 0, 0), (9, 0, 0), (8, 1, 0), (1, 8, 9)\}$

I do not understand how to "delete linearly independent vectors" from the span. Please help! Thanks.

Answer 1

You can see pretty easily that $S = \Bbb R^3$. To do this, let $(x,y,z) \in \Bbb R^3$. Then you can find $a,b,c,d \in \Bbb R$, such that $$(x,y,z) = a\cdot (0,0,0) + b \cdot(9,0,0) + c\cdot (8,1,0) + d \cdot (1,8,9) \; $$

Now we know that $\dim(\Bbb R^3) = \dim(S) = 3$. However, in your definition of $S$, $S$ is generated by 4 vectors. This means that you can remove one of them and the generated space of the remaining three vectors is still $S = \Bbb R^3$. So which vector can you remove, such that every $(x,y,z) \in \Bbb R^3$ can still be written as a linear combination of the remaining three vectors?