In Exercises 1322 a set S is given Determine by inspection a

In Exercises 13-22, a set S is given. Determine by inspection a subset of S containing the fewest vectors that has the same span as S. {[1 -2 3], [-2 4 -6]} {[1 0 2], [3 -1 1]} {[-3 2 0], [1 6 0], [0 0 0]} {[0 0 1], [0 1 2], [1 2 3], [2 3 4]} {[2 -3 5], [4 -6 10], [1 0 2]}

Solution

To find the subset of S containing the fewest vectors that has the same span as S, we have to prove that the third vector is a linear combination of first two vectors.

So for that:

The above equation can be separately written as:

That is,

Therefore,

So the third vector is in the span of the first two.

But also,

These two vectors are also linearly independent. Therefore, the subset of S containing the fewest vectors that has the same span as S is: