S2

Question 1

Two vectors are linearly dependent if and only if they lie on a line through the origin.

(To check the statement “A if and only if B”, you have to check two things: “A if B” and “A only if B”, which is the same as “B if A”.)

Answer 1

True

IF: If u and v are on a line through the origin, then v = cu for some c ∈ R. So cu − v = 0 is a nontrivial linear relation between them, hence they are dependent.

ONLY IF: If u and v linearly dependent, then au + bv = 0 for some a, b not both zero. If a is not zero, then u=(^−b/_a)v; if b is not zero, then v=(^−a/_b)u.

Eitherway we get that the two vectors lie on a line through the origin.

Question 2

If S is a linearly dependent set, then each vector in S is a linear combination of the other vectors in S.

Answer 2

False

The vectors (1,0), (0,2), (0,1) are linearly dependent, but the third is not a linear combination of the first two.

Question 3

If {x,y} is linearly independent, and {x,y,z} is linearly dependent, then z is in Span{x, y}.

Answer 3

True

There must be a relation ax+by+cz = 0 with a, b, c not all zero. Since x and y are independent, there is no relation ax + by = 0. Hence c ≠ 0, so

z = −(^a/_c)x − (^b/_c)y, which means that z is in the span of x and y.

Question 4

A linear transformation T is one-to-one if and only if T(x) = 0 has a unique solution x.

Answer 4

True

This is Theorem 11 in Chapter 1 of the book.

If T is one-to-one, then T (x) = 0 has at most one solution, and we know that 0 is a solution (T(0) = 0 because T is linear), so 0 must be the unique solution of T(x) = 0.

On the other hand, suppose that T (x) = 0 has a unique solution, which must be 0. Then T must be one-to-one, because if not, there would be a ≠ **b such that T(a) = T(b), and then T(a−b) = T(a)−(b) = 0, so a−b** would be another solution of T (x) = 0.

Question 5

If a linear transformation T with matrix A is one-to-one, then the columns of A span the codomain of T.

Answer 5

False

Take for instance A.

It is in echelon form, so we can see that it is one-to-one. But clearly the columns do not span ℝ³.

In fact, the columns of A spanning the codomain of T is equivalent to T being onto, which we know is not the same as being one-to-one.