Abstract Algebra: Theory and Applications · Problem 20.15

Fields — Thomas Judson

Mathematics — Abstract Algebra · linear-transformations

Problem

Linear Transformations. Let

V

and

W

be vector spaces over a field

F

, of dimensions

m

and

n

, respectively. If

T : V \to W

is a map satisfying

T(u + v) = T(u) + T(v)

T(\alpha v) = \alpha T(v)

for all

\alpha \in F

and all

u, v \in V

, then

T

is called a linear transformation from

V

into

W

. (a) Prove that the kernel of

T

\ker(T) = \{v \in V : T(v) = \mathbf{0}\}

, is a subspace of

V

. The kernel of

T

is sometimes called the null space of

T

. (b) Prove that the range or range space of

T

R(V) = \{w \in W : T(v) = w \text{ for some } v \in V\}

, is a subspace of

W

. (c) Show that

T : V \to W

is injective if and only if

\ker(T) = \{\mathbf{0}\}

. (d) Let

\{v_1, \ldots, v_k\}

be a basis for the null space of

T

. We can extend this basis to be a basis

\{v_1, \ldots, v_k, v_{k+1}, \ldots, v_m\}

V

. Why? Prove that

\{T(v_{k+1}), \ldots, T(v_m)\}

is a basis for the range of

T

. Conclude that the range of

T

has dimension

m - k

. (e) Let

\dim V = \dim W

. Show that a linear transformation

T : V \to W

is injective if and only if it is surjective. Hint. (a) Let

u, v \in \ker(T)

and

\alpha \in F

. Then

T(u + v) = T(u) + T(v) = \mathbf{0}

T(\alpha v) = \alpha T(v) = \alpha \mathbf{0} = \mathbf{0}.

Hence,

u + v, \alpha v \in \ker(T)

, and

\ker(T)

is a subspace of

V

. (c) The statement that

T(u) = T(v)

is equivalent to

T(u - v) = T(u) - T(v) = \mathbf{0}

, which is true if and only if

u - v = 0

u = v