Exam 1

Question 1

Let {u1,…, um} be a set of vectors. The span of this set is denoted
span{u1, u2,…, um}
and is defined to be the set of all linear combinations
x1 u1 + x2u2 + ⋯ + xmum

where x1, x2,…, xm can be any real numbers.

Answer 1

Span

Question 2

Let {u1, u2,…, um} be a set of vectors. If the only solution to the vector equation
x1 u1 + x2u2 + ⋯ + xmum = 0
is the trivial solution x1 = x2 = ⋯ = xm = 0, then the set is linearly independent.

Answer 2

Linear Independence

Question 3

A function T : Rm → Rn is a linear transformation if for all vectors u and v in Rm and all scalars r we have
(a) T(u + v) = T(u) + T(v)

(b) T(ru) = rT(u)

Answer 3

Linear Transformation

Question 4

A linear transformation T : Rm → Rn is one-to-one if for every vector w in Rn there exists at most one vector u in Rm such that T(u) = w. Alternate definition: A linear transformation T isone-to-oneif T(u) = T(v) implies that u = v.

Answer 4

One-to-one

Question 5

A linear transformation T : Rm → Rn is onto if for every vector w in Rn there exists at least one vector u in Rm such that T(u) = w.

Answer 5

Onto

Question 6

A linear transformation T : Rm → Rn is onto if for every vector w in Rn there exists at least one vector u in Rm such that T(u) = w.

Answer 6

Invertible

Question 7

A linear transformation T : Rm → Rn is invertible if T is one-to-one and onto. When T is invertible, the inverse function T−1 : Rn → Rm is defined by
T−1(y) = x   if and only if   T(x) = y

Answer 7

Inverse linear transformation