Let u4 be a vector that is not a linear combination of u1u2u

Let u4 be a vector that is not a linear combination of {u1,u2,u3}.

Select the best statement.

A. We only know that span{u1,u2,u3} span{u1,u2,u3,u4}.

B. There is no obvious relationship between span{u1,u2,u3} and span{u1,u2,u3,u4} .

C. span{u1,u2,u3} span{u1,u2,u3,u4}.

D. span{u1,u2,u3}= span{u1,u2,u3,u4} when u4 a scalar multiple of one of {u1,u2,u3}.

E. none of the above

Solution

right answer is C.

EXPLANATION D IS FALSE.If {u1, u2, u3} is a set of vectors in R n , and span{u1, u2, u3} = span{u1, u2, u3, u4} then u4 must be a linear combination of {u1, u2, u3}. TRUE. If the spans are the same, because u4 is in the span of {u1, u2, u3, u4}, it must also be in the span of {u1, u2, u3}. But, this implies u4 is a linear combination of {u1, u2, u3}

If {u1, u2, u3} is a spanning set for R n , then {u1, u2, u3, u4} also spans R n (where all vectors have the same dimension). TRUE. The span of a set of vectors is the set of all linear combinations of those vectors, i.e. all vectors v such that the equation v = x1u1 + x2u2 + x3u3 has a solution. But, if this equation has a solution, say x1 = c1, x2 = c2, x3 = c3, then the equation v = x1u1 + x2u2 + x3u3 + x4u4 is guaranteed to have at least one solution (namely x1 = c1, x2 = c2, x3 = c3, x4 = 0). But, this shows that v is also in the span of {u1, u2, u3, u4}. (We can’t take away solutions by adding a vector to the set; we could only add more. In other words, adding a vector to the set can only increase the span.)