2&3 Flashcards
To show U is not a subspace, any one of the three following statements will suffice:
- 0 ∉ U
- U is not closed under addition
- U is not closed under scalar multiplication
Definition 3.1: Vector space
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.
Definition 3.2: Subspace
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 ).
Suppose V is a vector space. Then a non-empty subset W of V is a subspace if and only if:
- 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).
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
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).
The range of an m x n matrix is the subset
R(A) = {Ax: x ϵ Rn} of Rm.
That is, the range is the set of all vectors y ∈ Rm 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
For any m x n matrix A: (hint: subspaces)
- R(A) is a subspace of Rm
- N(A) is a subspace of Rn
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.
Suppose that V is a vector space and that v1, v2, …, vk ϵ V. The linear span of X = {v1, …, vk} is the set of all linear combinations of the vectors v1, …, vk, denoted by Lin {v1, v2, …, vk} or Lin(X).
That is, Lin {v1, v2, …, vk} = {αv1 + … + αkvk: α1, α2, …, αk ϵ R}.
If X = {v1, v2, …, vk} 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, …, vk.
This subspace is also known as the subspace spanned by the set X ={v1, v2, …, vk}, or, simply, as the span,
of {v1, v2, …, vk}.
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
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.
If A is an m x n matrix the
row space will be a subspace of
Rn
If A is an m x n matrix the
column space will be a subspace of
Rm
Let V be a vector space and v1, …, vk ∈ V.
Then v1, v2, …, vk form a linearly independent set or are linearly independent if and only if
α1v1 + α2v2 + … + αkvk = 0 ⇒ α1 = α2 = … = αk = 0
Let V be a vector space and v1, …, vk ∈ V.
Then v1, v2, …, vk form a linearly dependent set or are linearly dependent if and only if
there are real numbers α1 = α2 = … = αk not all zero, such that α1v1 + α2v2 +…+ αmvm = 0
A set of two vectors is linearly dependent if and only if
one vector is a scalar multiple of the other.
The (i; j) minor of A, denoted by Mij , is
the determinant of the
(n - 1) x (n - 1) matrix obtained by removing the ith row and jth column of A.
The (i; j) cofactor of a matrix A is
Cij = (-1)i+jMij
Note: The cofactor is just the minor with the sign corresponding to its + or - position in the matrix
Method for determining determinant of A using row operations
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.
For an m x n matrix A, the null space of A is the subset
N(A) = {x ∈ Rn : Ax = 0}
of Rn, where 0 = (0; 0; : : : ; 0)T is the all-0 vector of length m.
{x : Ax = b} = [hint: use null space]
{x : Ax = b} = {x0 + z : z ∈ N(A)}
For an m x n matrix A, the null space of A is the subset:
N(A) = {x ∈ Rn : Ax = 0}
of Rn, where 0 = (0; 0; : : : ; 0)T is the all-0 vector of length m.
If X = {v1, v2, …, vk} 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, …, vk.
Let V be a vector space and v1,…, vk ∈ V.
Then v1,…, vk form a linearly independent set or are linearly independent if and only if
α1v1 + α2v2 + … + αkvk = 0 → α1 = α2 = … = αk = 0 :
that is, if and only if no non-trivial linear combination of v1; v2,…, vk equals the zero vector.
Let V be a vector space and v1,…, vk ∈ V.
Then v1,…, vk form a linearly dependent set or are linearly dependent if and only if
there are real real numbers α1, α2, … αk, not all zero, such that α1v1 + α2v2 + … + αkvk = 0
- Suppose that v1, v2, …, vk ∈ Rn. Then the set {v1, v2, …, vk} is linearly independent if and only if the matrix
(v1, v2, …, vk) has rank
- The maximum size of a linearly independent set of vectors in Rn is
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
Let V be a vector space. Then the subset
B = {v1, v2, …, vn} of V is said to be a basis for (or of) V if:
B is a linearly independent set of vectors, and
V = Lin(B).
B = {v1, v2, …, vn} is a basis of V if and only if
v ∈ V is a unique
linear combination of {v1, v2, …, vn}.
If determinant of AB ≠ 0 then this is the case.
Definition of dimension
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.
Steps to find a basis for a linear span in Rn:
Suppose we are given k vectors x1, x2, …, xk in Rn, and we want to find a basis for the linear span Lin{x1, x2, …, xk}.
- Write a matrix A whose rows are the vectors x1, x2, …, xk
- Reduce it to echelon form by elementary row operations,
- Basis is formed by the rows (transposed to columns) of the echelon matrix with leading ones OR we can take the original xi that correspond to rows i that have leading ones.
As we have seen, the range and null space of an m x n matrix are […?] of Rm and Rn (respectively).
subspaces
The rank of a matrix A is …
and the nullity is …
rank (A) = dim (R(A))
nullity (A) = dim (N(A))
Steps to find a basis for the column space of A
- Row reduce A to echelon form
- Basis formed by columns of A (not of echelon matrix!!!) that correspond to position of leading ones in rref(A)
For an m x n matrix A, rank(A) + nullity(A)
rank(A) + nullity(A) = n
Steps to find a basis for the row space of A
- Row reduce A to echelon form
- 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)
Steps to find a basis for the null space of A
- Row reduce A to echelon form
- Set up rref(A)x = 0
- Find general solution in terms of fixed and free variables and form basis from that.
