Linear Algebra: Vector Spaces, Subspaces, and Basis Concepts

V = R^n is a vector space

True

V = {(x, y )|x, y ∈ R} = R^2 is a vector space

True

V = {(x, y , z)|x, y , z ∈ R} = R^3 is a vector space

True

If we take away "or less" and just have polynomials of

degree n

Then we don't have a vector space

The m × n matrices, Mmn, form a vector space under the

operations

Matrix addition and scalar multiplication

Is the set of 2 × 2 matrices of the form

[a 0

b c]

a vector space? If not, what does it defy?

Yes

Is the set V = {(x, 1)|x ∈ R} with the usual addition a vector space? If not, what does it defy?

No; Not closed under addition

Is C [0, 1] - the set of continuous functions on a closed interval

under the following operations.

(f + g )(x) = f (x) + g (x)

(cf )(x) = cf (x) a vector space? If not, what does it defy?

Yes

Vector spaces to know about

Rn

Pn, the set of all polynomials of degree n or less

P, set of all polynomials

Mmn, the set of all m × n matrices

C [a, b], continuous functions on a closed interval

C (−∞, ∞), continuous functions on the real line.

Is V =

{[a b

0 1]|a, b, c ∈ R}

with usual operations a vector space?

No; Not closed under addition

A nonempty subset W of a vector space V is called a

subspace of V if W is itself a vector space under the operations

of addition and scalar multiplication defined in V.

Is the set of 2 × 2 nonsingular (invertible) matrices a subspace

of M22?

No, invertible = is never a subspace.

Is W = {(x, 3)|x ∈ R} a subspace of R^2?

No, must have (0,0) as the zero vector.

Subspaces of R^3

R3 is a subspace (trivially).

{~0} = {(0, 0, 0)} is the trivial subspace.

Any line through the origin is a subspace.

Any plane through the origin is a subspace.

Let S = {~v1, ~v2, . . . , ~vn} be a subset of a vector space V . S

is a spanning set of V if

Every vector in V can be written as a

linear combination of vectors in S.

S = {(1, 0), (0, 1)} is a spanning set for R2.

True

S =

{[1 0

0 0],

[0 1

0 0],

[0 0

1 0],

[0 0

0 1]}

is a spanning set of M22.

True

S = {1, x, x2, x3} is a spanning set for P3, all polynomials of

degree 3 or less.

True

S = {(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1)} is a

spanning set for Rn.

True

s S = {(5, 0), (5, −4)} a spanning set for R2?

Literally just find the determinant... on the diagonal minus the off-diagonal for 2x2 and into the calculator for any other. If 0, not a spanning set.

How to calculate a spanning set

Determinent

How to calculate linear independence

RREF on calc

A set of vectors S = {~v1, . . . , ~vk } in V is a basis for V if

1 S spans V

2 S is linearly independent.

True

The dimension of a vector space V, dim V, is the number of

elements in the basis for V.

Ex. B = {(1, 0), (0, 1)} is a basis for R2. dim R2 = 2

Ex. B = {(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1)} is a basis

for Rn, called the standard basis. dim Rn = n

Is S = {(5, 0), (5, −4)} a basis for R2?

The set S is (1) spanning and (2) linearly independent.

S is a basis for R^2

Theorem. If dim V = n, i.e. with a basis {~v1, ~v2, . . . , ~vn}, then

Every set containing more than n vectors is linearly dependent.

every set containing less than n vectors is not a spanning set.

Rank of a matrix

Number of non-zero rows

Basis for a row space

The non-zero rows of the RREF form a basis for the row space.

Basis for a column space

Write the columns of the original matrix as the rows of a new matrix, RREF in the calculator, take rows and use them as columns

Dimension of a row space

Number of linearly independent rows, same as / equal to rank

Rank

Number of linearly independent rows

Dimension of a column space

Equal to rank and row space dimension

n

number of columns

Formula for n

Rank + Nullity

Dimension of a solution space

Dimension of a null space

If ~x can be written as ~x = c1~v1 + c2~v2 + · · · + cn ~vn, then ~x is

called a

Linear Combination

Example of a linear combination of vectors

(2, 1, 3)= 2(1, 0, 0) + 1(0, 1, 0) + 3(0, 0, 1).

Let V be a set with two operations of vector addition and

scalar multiplication. If 1-10 are satisfied for every ~u, ~v, ~w in V

and every scalar c and d, then V is a

Vector Space

Vector space

A vector space is a set of "vectors", a set of scalars, and

two operations. The set of scalars is R for us.

Find a basis for W = {(5t, −3t, t, t)|t ∈ R} and determine the

dimension of the subspace W of R4

5t, −3t, t, t) = t(5, −3, 1, 1).

So W is spanned by S = {(5, −3, 1, 1)}.

Clearly, S is linearly independent. So, dim W = 1

Let A be an m × n matrix. To get the basis of the row space of A:

1 Write A in row echelon form B.

2 The basis is formed by the nonzero row vectors of B

How to find the basis of a column space:

1 Write A in row echelon form.

2 Find columns with leading 1's.

3 Then choose those columns from the original matrix A. They form a basis

The dimension of the row space equals the dimension

of the column space.

True

Rank

The dimension of the row space of A is called the rank of A

Dimension

Column or rank basis space (number of nonempty rows or columns)

Formula for n

rank(A) + nullity(A) = n the number of columns of A

Rank + Nullity Theorem

Let A be an m × n matrix.

If rank(A) = r , then the dimension of the nullspace if n − r

Process for finding the basis of a nullspace

1. RREF

2. Ax = 0 for all rows, set parameters t or s and t

3. solve for x1 and x2 or x1, x2 and x3

4. plus in t and s

5. Factor out t and s, get a basis for the nullspace, and makesure to add ()^T

Example of finding nullspace

Find the basis for the nullspace of A =

[ 3 −6 2|01

−2 4 −14|0

1 −2 7|0]

RREF

−→

[ 1 −2 7|0

0 0 0|0

0 0 0 |0]

x3 = t for any real t

x2 = s for any real s

x1 − 2x2 + 7x3 = 0, x1 = 2s − 7t

The nullspace is W = {(2s − 7t, 1s + 0t, 0s + 1t)T |s, t ∈ R}

The basis for the nullspace is {(2, 1, 0)T, (−7, 0, 1)T }.

dim(N(A)) = 2.

Transition matrix theorem

P is the transition matrix from a basis B′ to a basis

B in Rn, then P is invertible and the transition matrix from B to

B′ is given by P−1

Find B to B'

Actually solve by doing B' to B