chapter 4

Question 1

Let U and V be two vector spaces. A linear map (or linear transformation)
between U and V is a map T :U →V which..

Answer 1

..preserves vector space operations

Question 2

linear maps can be formed by combining other linear maps via

Answer 2

addition, composition, scalar multiplication

Question 3

Let U be a vector space with basis (e

1,…,en). A linear map T : U → V is completely determined by

Answer 3

by the images of the basis vectors T (e1),…,T(en).

Question 4

The sum of two linear maps T1,T2 :U →V is defined as;

The multiplication of a linear map T :U →V by a scalar λ ∈ R is defined as

Answer 4

(T1 +T2)(x) = T1(x)+T2(x) for every x ∈U,

(λT )(x) = λ.T (x) for every x ∈U.

Question 5

we have the standard ‘rules of matrix arithmetic’. Whenever the operations are welldefined, we have:

Answer 5

• A +B = B + A,
• A +(B +C) = (A +B)+C
• A(B +C) = AB + AC,
• (A +B)C = AC = BC,
• λ(AB) = (λA)B = A(λB).

Question 6

The transpose of an m ×n matrix A is an n×m matrix denoted AT such that

Answer 6

(AT)ij = Aji

Question 7

A matrix A ∈ Mn(R) is called invertible iff

Answer 7

there exists another matrix B ∈ Mn(R)

such that AB = In. The matrix B is called the inverse of A

Question 8

row vectors/column vectors of matrices

Answer 8

row vectors of A - the subspace they spanned is called the row space of A and its dimension, denoted rkr (A), is called the row rank of A.
column vectors of A - the subspace they spanned is called the column space of A
and its dimension, denoted rkc (A), is called the column rank of A.

Question 9

Let A be an m×n matrix. The subspace of solution of the homogeneous system
of linear equations Ax = 0 has dimension

Answer 9

n −rk(A).

Question 10

Let A be an n ×n matrix, Then:

A is invertible ⇐⇒

Answer 10

⇐⇒ rk(A) = n.

Question 11

Let T : U → V be a linear map. The kernel (or null space) of T , denoted by N(T) or ker(T) is defined as

Answer 11

N(T)= {u∈U | T(u)=0V}.

Question 12

The range space of T , denoted by R(T ) is defined as the image of U under T :

Answer 12

R(T)={v∈V | v=T(u) for some u∈U} = {T(u) | u∈U}.

Question 13

The nullspace of a linear map T : U → V is a vector subspace of U and the range is

Answer 13

a vector subspace of V .

Question 14

Let A be an m×n matrix and let T:Rn→Rm be the linear map T(x):=Ax. Then:

Answer 14

(i) the null space of T is the solution space to the homogeneous system of linear equations Ax = 0,
(ii) the range of T is the column space of A.

Question 15

If T : U → V is a linear map, then the dimension of the range of T is called the (1). The dimension of the kernel of T is called (2)

Answer 15

(1) rank of T: rk(T):=dimR(T).

(2) the nullity of T.

Question 16

rank-nullity theorem

Answer 16

If T : U → V is a linear map between finite-dimensional vector spaces, then we have:
dimU =dimN(T)+dimR(T).

Question 17

Recall that a map f :X →Y between sets X and Y is called:

surjective if, injective if, bijective if

Answer 17

(i) surjective if for every y ∈ Y , there exists at least one point x ∈ X such that f (x) = y.
(ii) injective if for every y ∈ Y , there exists at most one point x ∈ X such that f (x) = y. In other words, for every x, x′ ∈ X , f (x) = f (x′) ⇒ x = x′.
(iii) bijective if it is both injective and surjective. In other words, for every y∈Y, there exists exactly one point x∈X such that f(x)=y.

Question 18

Let T : U → V be a linear map between two vector spaces. Then we have:

Answer 18

(i) T is surjective ⇔ rk(T ) = dim(V ).

(ii) T is injective⇔N(T)={0U}.

Question 19

Let T : U → V be a linear map between two vector spaces of the same dimension. Then:

Answer 19

T is an isomorphism ⇔ T is injective ⇔ T is surjective.