Chapter 4: Basis, dimension and rank

Question 1

Definition A subspace of Rn
is a set U of vectors in Rn
such that:

Answer 1

(1) The zero vector 0 ∈ U.
(2) If x and y are in U then x + y is in U.
(3) If x is in U and c ∈ R is a scalar, then cx ∈ U.

Question 2

Definition Let A be an m × n matrix.
a) The null space of A, written null(A), is

Answer 2

the subspace of solutions to the homogeneous equation Ax = 0. Thus
null(A) = {x ∈ R^n: Ax = 0}.

Question 3

Definition Let A be an m × n matrix.
b) The image space of A, written im(A) is

Answer 3

the set of all vectors
y ∈ Rm such that Ax = y has a solution. Thus
im(A) = {Ax : x ∈ R^n}.

Question 4

Definition Let A be an n × n matrix and let λ be an eigenvalue of A. The λ-eigenspace of A and is
denoted E_λ is

Answer 4

The collection of all eigenvectors corresponding to λ together with the zero vector.
Equivalently
Eλ = {x ∈ R^n: Ax = λx}.

Question 5

Definition. If V = {v1, v2, · · · , vk} is a set of vectors in R^n, then the span of V is

Answer 5

then the set of all linear combinations of v1, v2, · · · , vk. Symbolically,
span V = {∑^k_i=1 vici: c1, c2, · · · , ck ∈ R}.

Question 6

Theorem 5.1.1 A span is a subspace Let v1, . . . , vk be vectors in R^n. Then

Answer 6

a) U = span{v1, . . . , vk} is a subspace of Rn.
b) If W is a subspace of Rn and each vi ∈ W, then U ⊆ W.

Question 7

Definition A set {v1, v2, · · · , vk} of vectors in R^n
is linearly independent if and only if

Answer 7

the only solution to the equation
c1v1 + c2v2 + · · · + ckvk = 0, ci∈ R
is c1 = c2 = · · · = ck = 0

Question 8

Theorem 5.2.2 If A is an m × n matrix, let a1, · · · , an. denote the columns of A.

Answer 8

a) {a1, · · · , an} is independent in Rm if and only if Ax = 0, x ∈ Rn, implies x = 0.
b) Rn = span{a1, · · · , an} if and only if Ax = b has a solution x for every vector b ∈ Rm.

Question 9

Theorem 5.2.3 The following are equivalent for an n × n matrix
A.

Answer 9

a) A is invertible.
b) The columns of A are linearly independent.
c) The columns of A span Rn
d) The rows of A are linearly independent.
e) The rows of A span the set of all 1 × n rows.

Question 10

Definition A basis for a subspace S of Rn
is a set of vectors in S that

Answer 10

a) spans S, and
b) is linearly independent.

Question 11

Theorem 5.2.5 The basis theorem

Answer 11

Let S be a subspace of Rn.
Then any two bases for S have the same number of vectors.

Question 12

Dimension

Answer 12

If S is a nontrivial subspace of Rn
, then the number
of vectors in a basis for S is called the dimension of S, denoted
dim(S). If S = {0} then we define dim(S) = 0.

Question 13

Theorem 5.2.6 Let U =/= {0} be a subspace of Rn. Then

Answer 13

a) U has a basis and dim U ≤ n.
b) Any independent set in U can be enlarged (by adding vectors
from any fixed basis of U) to a basis of U, if not already so.
c) Any spanning set for U can be cut down (by deleting vectors)
to a basis of U, if not already so.

Question 14

Row space of matrix A, an m × n matrix.

Answer 14

written row(A) is the subspace of Rn
spanned by the rows of A.

Question 15

Column space of matrix A, an m × n matrix.

Answer 15

written col(A), is the subspace of Rm
spanned by the columns of A.

Question 16

Lemma 5.4.1 Let and A B denote m × n matrices.

Answer 16

a) If A → B by elementary row operations, then row(A) = row(B).
b) If A → B by elementary column operations, then col(A) = col(B).

Question 17

Lemma 5.4.2 If R is a row-echelon matrix, then

Answer 17

a) The nonzero rows of R are a basis of row(R).
b) The columns of R containing leading ones are a basis of col(R).

Question 18

rank(A)=

Answer 18

dim(row(A))=dim(col(A))

Question 19

Theorem 5.4.1 Rank Theorem Let A be any m × n matrix of
rank r. Then

Answer 19

dim(row(A))= dim(col(A)).
Moreover, if A is carried to a row-echelon matrix R by row operations, then
a) The r nonzero rows of R are a basis of row A.
b) If the leading 1s lie in columns j1, j2, . . . , jr of R, then columns
j1, j2, . . . , jr of A are a basis of col(A).

Question 20

Rank of a matrix A

Answer 20

the dimension of its row and column spaces and is denoted rank(A).

Question 21

Corollary 5.4.1 For any matrix A, rank(A) =

Answer 21

rank(A^T).

Question 22

Definition The nullity of a matrix A

Answer 22

the dimension of its null space null(A), and is denoted nullity(A).

Question 23

Theorem 5.4.2 The rank and nullity theorem If A is an m × n matrix then

Answer 23

rank(A) + nullity(A) = n.

Question 24

Definition Let A and B be n × n matrices. We say that A is similar to B if

Answer 24

there exists an invertible n × n matrix P such that P^−1AP = B. If A is similar to B we write A ∼ B, where ∼ denotes an equivalence relation.

Question 25

Theorem 5.5.1 Let A and B be n × n matrices with A ∼ B. Then:

Answer 25

(a) det(A) = det(B);
(b) A is invertible if and only if B is invertible;
(c) A and B have the same rank;
(d) A and B have the same characteristic polynomial;
(e) A and B have the same eigenvalues.
(f) A and B have the same trace.

Question 26

Definition An n × n matrix A is diagonalisable if

Answer 26

it is similar to a diagonal matrix.

Question 27

Theorem 5.5.3 Let A be an n × n matrix. Then A is diagonalisable if and only if it

Answer 27

has n linearly independent eigenvectors.
Moreover, there exists an invertible matrix P and a diagonal matrix D such that P^−1AP = D if and only if:
* The columns of P are n linearly independent eigenvectors of A;
* The diagonal entries of D are the eigenvalues of A corresponding to the eigenvectors in P (in the same order).

Question 28

Theorem 5.5.4 and 5.5.5 Let A be an n × n matrix with n distinct eigenvalues with corresponding eigenvectors x1, . . . , xn .
Then

Answer 28

{x1, . . . , xn} is a linearly independent set and A is diagonalisable.

Question 29

Algebraiic multiplicity of an eigenvalue

Answer 29

its multiplicity as a root of the characteristic equation. The geometric
multiplicity of an eigenvalue λ is the dimension of its eigenspace
Eλ.

Question 30

Theorem 5.5.6 (The diagonalisation theorem) Let A be an n × n matrix whose distinct eigenvalues are λ1, . . . , λk. The following are equivalent:

Answer 30

(a) A is diagonalisable;
(b) The union of a basis for the each of the eigenspaces of A contains n vectors;
(c) The algebraic multiplicity of each eigenvalue is equal to itsvgeometric multiplicity and the sum of these multiplicities across all eigenvalues is n.

Question 31

Theorem 9.2.2 Let B = {u1, . . . , un} and C = {v1, . . . , vn} be ordered bases for Rn and let PC←B be the change of basis matrix from B to C. Then for all x ∈ Rn

Answer 31

a) PC←B[x]B = [x]C;
b) PC←B is the unique n × n matrix P such that P[x]B = [x]C
c) PC←B is invertible and (PC←B)^−1 = PB←C

Question 32

Theorem Gauss Jordan method for computing a change of
basis matrix

Answer 32

[C|B] → [In|PC←B]