2&3

Question 1

To show U is not a subspace, any one of the three following statements will suffice:

Answer 1

1. 0 ∉ U
2. U is not closed under addition
3. U is not closed under scalar multiplication

Question 2

Definition 3.1: Vector space

Answer 2

A vector space V is a set equipped with an addition
operation and a scalar multiplication operation such that for all α, β ϵ R and all
u, v, w ϵ V ,
1. u+ v ϵ V (closure under addition)
2. u + v = v + u (the commutative law for addition)
3. u + (v + w) = (u + v) + w (the associative law for addition)
4. there is a single member 0 of V, called the zero vector, such that for all v ϵ V ,
v + 0 = v
5. for every v ϵ V there is an element w ϵ V (usually written as -v), called the
negative of v, such that v + w = 0
6. αv ϵ V (closure under scalar multiplication)
7. α(u + v) = αu + αv (distributive under scalar multiplication)
8. (α + β )v = αv + βv
9. α(βv) = (αβ)v
10. 1v = v.

Question 3

Definition 3.2: Subspace

Answer 3

A subspace W of a vector space V is a non-empty subset of V that is itself a vector space (under the same operations of addition and scalar
multiplication as V ).

Question 4

Suppose V is a vector space. Then a non-empty subset W of V is a subspace if and only if:

Answer 4

• for all u; v ϵ W, u + v ϵ W (W is closed under addition), and
• for all v ϵ W and α ϵ R, αv ϵ W (W is closed under scalar multiplication).

Question 5

Generally, if W is a subspace of a vector space V and

x ϵ V, the set x + W defined by x + W = {x + w: w ϵ W}

is called an

Answer 5

affine subset of V. An affine subset is not generally a subspace (although
every subspace is an affine subset, as we can see by taking x = 0).

Question 6

The range of an m x n matrix is the subset

Answer 6

R(A) = {Ax: x ϵ Rⁿ} of R^m.

That is, the range is the set of all vectors y ∈ R^m of the form y = Ax for some x ∈ Rⁿ.

The range of A can be described as the set of all linear combinations of the columns of A

Question 7

For any m x n matrix A: (hint: subspaces)

Answer 7

• R(A) is a subspace of R^m
• N(A) is a subspace of Rⁿ

Question 8

The set of all linear combinations of a given set of vectors of a vector space V forms a
subspace, and we give it a special name.

Answer 8

Suppose that V is a vector space and that v₁, v₂, …, v_k ϵ V. The linear span of X = {v₁, …, v_k} is the set of all linear combinations of the vectors v_{1, …}, v_k, denoted by Lin {v₁, v₂, …, v_k} or Lin(X).

That is, Lin {v₁, v₂, …, v_k} = {αv₁ + … + α_kv_k: α₁, α₂, …, α_k ϵ R}.

If X = {v₁, v₂, …, v_k} is a set of vectors of a vector space V, then Lin(X) is a subspace of V. It is the smallest subspace containing the vectors v₁, v₂, …, v_k.

This subspace is also known as the subspace spanned by the set X ={v₁, v₂, …, v_k}, or, simply, as the span,
of {v₁, v₂, …, v_k}.

Question 9

In R³ a plane is defined as the set of all vectors x whose components satisfy a single Cartesian equation,

ax + by + cz = d.1

If d = 0 then

Answer 9

the plane contains the origin, and the position vectors of points on the plane form a subspace of R³.

If S is the linear span of two vectors, it is a subspace of R³.

If d ≠ 0 then the plane is not a subspace. It is an affine set, a translation of a linear space.

Question 10

If A is an m x n matrix the
row space will be a subspace of

Answer 10

Rⁿ

Question 11

If A is an m x n matrix the
column space will be a subspace of

Answer 11

R^m

Question 12

Let V be a vector space and v₁, …, v_k ∈ V.
Then v₁, v₂, …, v_k form a linearly independent set or are linearly independent if and only if

Answer 12

α₁v₁ + α₂v₂ + … + α_kv_k = 0 ⇒ α₁ = α₂ = … = α_k = 0

Question 13

Let V be a vector space and v₁, …, v_k ∈ V.
Then v₁, v₂, …, v_k form a linearly dependent set or are linearly dependent if and only if

Answer 13

there are real numbers α₁ = α₂ = … = α_k not all zero, such that α₁v₁ + α₂v₂ +…+ α_mv_m = 0

A set of two vectors is linearly dependent if and only if
one vector is a scalar multiple of the other.

Question 14

The (i; j) minor of A, denoted by M_ij , is

Answer 14

the determinant of the
(n - 1) x (n - 1) matrix obtained by removing the ith row and jth column of A.

Question 15

The (i; j) cofactor of a matrix A is

Answer 15

C_ij = (-1)^i+jM_ij

Note: The cofactor is just the minor with the sign corresponding to its + or - position in the matrix

Question 16

Method for determining determinant of A using row operations

Answer 16

Reduce matrix A to an upper triangle matrix.

Call it T.

IAI is ITI plus whatever shit you’ve done to it: minus if you swapped rows, multiple if you multiplied a row.

Question 17

For an m x n matrix A, the null space of A is the subset

Answer 17

N(A) = {x ∈ Rⁿ : Ax = 0}
of Rⁿ, where 0 = (0; 0; : : : ; 0)^T is the all-0 vector of length m.

Question 18

{x : Ax = b} = [hint: use null space]

Answer 18

{x : Ax = b} = {x₀ + z : z ∈ N(A)}

Question 19

For an m x n matrix A, the null space of A is the subset:

Answer 19

N(A) = {x ∈ Rⁿ : Ax = 0}
of Rⁿ, where 0 = (0; 0; : : : ; 0)^T is the all-0 vector of length m.

Question 20

If X = {v₁, v₂, …, v_k} is a set of vectors of a vector space V, then Lin(X) is

Answer 20

a subspace of V. It is the smallest subspace containing the vectors v1, v2, …, vk.

Question 21

Let V be a vector space and v_1,…, v_k ∈ V.
Then v_1,…, v_k form a linearly independent set or are linearly independent if and only if

Answer 21

α₁v₁ + α₂v₂ + … + α_kv_k = 0 → α₁ = α₂ = … = α_k = 0 :
that is, if and only if no non-trivial linear combination of v₁; v₂,…, v_k equals the zero vector.

Question 22

Let V be a vector space and v₁,…, v_k ∈ V.
Then v₁,…, v_k form a linearly dependent set or are linearly dependent if and only if

Answer 22

there are real real numbers α₁, α₂, … α_k, not all zero, such that α₁v₁ + α₂v₂ + … + α_kv_k = 0

Question 23

1. Suppose that v₁, v₂, …, v_k ∈ Rⁿ. Then the set {v₁, v₂, …, v_k} is linearly independent if and only if the matrix

(v₁, v₂, …, v_k) has rank

2. The maximum size of a linearly independent set of vectors in Rⁿ is

Answer 23

k (vectors are linearly indep iff rank of rref(A) has k leading ones, i.e. has rank k, has rank same as number of vectors)

n

Question 24

Let V be a vector space. Then the subset
B = {v₁, v₂, …, v_n} of V is said to be a basis for (or of) V if:

Answer 24

B is a linearly independent set of vectors, and
V = Lin(B).

Question 25

B = {v₁, v₂, …, v_n} is a basis of V if and only if

Answer 25

v ∈ V is a unique
linear combination of {v₁, v₂, …, v_n}.

If determinant of A_B ≠ 0 then this is the case.

Question 26

Definition of dimension

Answer 26

The number d of vectors in a finite basis of a vector
space V is the dimension of V and is denoted dim(V). The vector space V = {0} is defined to have dimension 0.

Question 27

Steps to find a basis for a linear span in Rⁿ:

Suppose we are given k vectors x_1, x_2, …, x_k in Rⁿ, and we want to find a basis for the linear span Lin{x_1, x_2, …, x_k}.

Answer 27

1. Write a matrix A whose rows are the vectors x_1, x_2, …, x_k
2. Reduce it to echelon form by elementary row operations,
3. Basis is formed by the rows (transposed to columns) of the echelon matrix with leading ones OR we can take the original x_i that correspond to rows i that have leading ones.

Question 28

As we have seen, the range and null space of an m x n matrix are […?] of R^m and Rⁿ (respectively).

Answer 28

subspaces

Question 29

The rank of a matrix A is …

and the nullity is …

Answer 29

rank (A) = dim (R(A))

nullity (A) = dim (N(A))

Question 30

Steps to find a basis for the column space of A

Answer 30

1. Row reduce A to echelon form
2. Basis formed by columns of A (not of echelon matrix!!!) that correspond to position of leading ones in rref(A)

Question 31

For an m x n matrix A, rank(A) + nullity(A)

Answer 31

rank(A) + nullity(A) = n

Question 32

Steps to find a basis for the row space of A

Answer 32

1. Row reduce A to echelon form
2. Basis formed by either non-zero rows (transposed) of rref(A) or rows of A that correspond to position of leading ones in rref(A)

Question 33

Steps to find a basis for the null space of A

Answer 33

1. Row reduce A to echelon form
2. Set up rref(A)x = 0
3. Find general solution in terms of fixed and free variables and form basis from that.