Question

Let v1 = -4
-1
-2

v2 = -3
1
-2

v3= 1
-5
2

and H = Span{v1,v2,v3} . Note that v3 = 2v1 - 3v2.

Which of the following sets form a basis for the subspace H, i.e., which sets form an efficient spanning set containing no unnecessary vectors?


a. {V1, V2, V3}
b. {V1, V2}
c. {V1,V3}
d. {V2,V3}

Answer 1

We're told (and we can confirm) that $v_3=2v_1-3v_2$, so $v_3$ is a linear combination of the other two vectors.

This means H is sufficiently spanned by $\{v_1,v_2\}$; no need for the third vector.

But this also means we can write either $v_1$ as a linear combination of $\{v_2,v_3\}$, and $v_2$ as a lin. com. of $\{v_1,v_3\}$. So any set of these three vectors taken two at a time will span the subspace H. Hence all of b, c, and d are acceptable.