Linear Transformations

Question 1

Linear Transformation

Synonyms

Answer 1

• linear mapping
• linear map
• linear operator
• homomorphism

Question 2

Linear Transformation

Definition

Answer 2

• let V, W be vector spaces over F, a map T:V->W is called a linear transformation if:
  i) T(|v + |v’) = T(|v) + T(|v’) , for all v,v’ϵV
  ii) T(a|v) = aT(|v) , for all aϵF and vϵV

Question 3

Elementary Properties of Linear Transformations

Answer 3

-if T is a linear mapping then:
1) T(|0) = |0, if T(|0) ≠|0 then T is not linear
2) T(|-v) = -T(|v)
3) T(a|v + b|v’) = aT(|v) + bT(|v’), linear transformations preserve linear combinarions
-more generally, if {v1,v2,…,vn}ϵV and {a1,a2,…,an}ϵF
T(Σai |vi ) = Σ[ai T(|vi) ]
where sums are between i=1 and i=n

Question 4

Linear Transformations

Special Cases

Answer 4

i) The Identity Map: Idv : |vϵV -> |vϵV is a linear map V->W
ii) The Zero Map: Z : |vϵV -> |0ϵW
iii) mxn Matrix A Ta : |vϵF^n -> A|vϵF^m
for |v={v1,v2,…,vn}ϵF^n :
Ta(|v) = A |v

Question 5

Abstract Examples of Linear Transformations

Answer 5

1) Differentiation
2) Integration
3) The Definite Integral
4) Transposition Maps (transforms a matrix to its transpose
5) Trace Map (gives the trace of a matrix, the sum of all elements in the main diagonal

Question 6

Linear Transformations

Matrix Proposition

Answer 6

-given a linear map T:F^n->F^m, there exists a uniquely defined matrix A such that T=Ta:|vϵF->A |v ϵF^m

Question 7

List of Example Linear Transformations in ℝ² (and ℝ^3)

Answer 7

• Rotation
• Reflection
• Rescalings (Enlargement)

Question 8

Transformation Matrix for Rotation in ℝ² (and ℝ^3)

Answer 8

A = (cosθ -sinθ)

(sinθ cosθ)

Question 9

Matrix of Linear Transformation

Answer 9

-let V, W be finite dimensional vector spaces
-linear maps T:F^n -> F^m are always given by matrices
-there exists a unique matrix A with T=Ta: |vϵF^n -> A|vϵF^m
-the same is true for linear maps T: V->W, but bases must be fixed for V & W, the matrices we get depend on these bases
-let T: V->W be a linear map
let S={v1,…,vn} be a basis of V, let R={w1,…,wn} be a basis of W
-the matrix of T with respect to the bases is the matrix
A=aij whose jth column is T(|vj)
so
T(|v1) = a11 |w1 + a21 |w2 + … + am1 |wm
T(|v2) = a12 |w1 + a22 |w2 + … + am2 |wm
….
T(|vn) = a1n |w1 + a2n |w2 + … + amn |wm

Question 10

How do you find the matrix of a transformation T with respect to the canonical bases of V and W?

Answer 10

• take the vectors that form the basis of V
• apply the transformation T to them
• write this results as coordinate vectors using the basis of W
• these coordinate vectors form the columns of the matrix A

Question 11

How do you apply a transformation using the matrix of transformation with respect to the canonical bases?

Answer 11

• let T: V -> W
• write the vector as coordinate vectors in terms of the basis of V
• multiply the coordinate vectors by the transformation matrix, A
• the resulting vector will give you the coordinate vector with respect to the basis of W

Question 12

Kernel

Definition

Answer 12

-let T: V->W be a linear mapping between vector spaces
-the null space or kernel of T is:
ker T = {vϵV : T(v)=0w} ⊆ V
-the kernel of T is the vector space of all vectors v in V which become the zero vector in W when the transformation T is applied to them

Question 13

Image

Definition

Answer 13

-let T: V->W be a linear mapping between vector spaces
-the image of T is:
im T = {T(v) : vϵV} ⊆ W
-the image of T is the vector space in W of all vectors that result from the transformation T being applied to all vectors v in V

Question 14

Kernel

Matrix Definition

Answer 14

-if A is an mxn matrix and Ta : vϵF^n -> AvϵF^m , then:
ker Ta = {vϵF^n : Av=|0}
i.e. the kernel of A is the solution space for the system of linear equations given by A

Question 15

Image

Matrix Definition

Answer 15

-if A is an mxn matrix and Ta : vϵF^n -> AvϵF^m , then:
im Ta = {Av : vϵF^n} = bϵF^m
i.e. the system of linear equations Av = b is consistent/has solutions

Question 16

Kernel, Image and Subspace

Answer 16

• the kernel and image are both subspaces
• let T: V -> W be a linear map
• the kernel of T is a subspace of V and the image of T is a subspace of W

Question 17

Nullity

Definition

Answer 17

-the nullity of T is dim (ker T), written:

n(T) = dim( ker(T) )

Question 18

Rank

Definition

Answer 18

-the rank of T is dim(im(T)), written:

r(T) = dim( im(T) )

Question 19

What are the source and target?

Answer 19

• the source is another name for the domain

- the target is the same as the codomain

Question 20

Composition of Linear Maps

Answer 20

-if S:U->V and T:V->W are linear maps, then the composition:

T∘ S : U->W is linear too

Question 21

Composition of Linear Maps with Matrices

Answer 21

-given an mxn matrix A and an nxl matrix B we have linear maps
Tb : F^l -> F^n where Tb(v)=Bv and Ta : F^n->F^m where Ta(v)=Av
-the composition is Ta∘ Tb = Tab
i.e. (Ta∘Tb)(v) = ABv

Question 22

Transition Matrices From One Basis to Another

Answer 22

• let U, V and W be vector spaces, T:U->V and T’:V->W are linear maps
• consider bases S of U, S’ of V and S’’ of W
• let A(S’,S) be the matrix of T w.r.t. S and S’
• let A’(S’‘,S’) be the matrix of T’ w.r.t. S’ and S’’
• then with respect to S & S’’ the matrix B=B(S’‘,S) of (T’∘T) : U->W is B(S’‘,S) = A’(S’‘,S’) A(S’,S)

Question 23

Transition Matrices - Identity Transformation

Answer 23

-if id:V->V is the identity map and S, S’ are bases of V
-the transition matrix, writing S’ in terms of S, is the matrix I(S,S’) of id with respect to S’ in the domain of id and S in the codomain of id
I(S,S’) = (I(S’,S))^(-1)

Question 24

Transition Matrix With Respect to Other Matrices

Derivation

Answer 24

-let T:V->W be a linear map
-S and S’ are bases of V & R and R’ are bases of W
-if A=A(R,S) is the matrix of T with respect to S and R
and A’=A(R’,S’), then:
A(R’,S’) = I(R’,R) A(R,S) I(S,S’)
i.e. A’ = Q^(-1) A P

Question 25

Transition Matrix With Respect to Other Matrices

Equation

Answer 25

A’ = Q^(-1) A P

A’ = matrix of T with respect to R’ and S’
A = matrix of T with respect to R and S
P = I(S,S’) matrix of id:V->V w.r.t. S’ in domain and S in codomain (transition matrix writing S in terms of S’)
Q = I(R,R’) matrix od id:W->W w.r.t. R’ in domain and R in the codomain (transition matrix writing R’ in terms of R)
-> Q^(-1) = I(R’R) matrix of id:W->W w.r.t. R in domain and R’ in the codomain

Question 26

Equivalence

Definition

Answer 26

-two mxn matrices A and A’ are equivalent if there are non-singular (invertible) matrices P and Q such that A’=Q^(-1) A P

Question 27

Similar

Definition

Answer 27

-two nxn matrices B and B’ are similar if there is a non-singular matrix P such that B’ = P^(-1) B P

Question 28

Equivalence Proposition

Answer 28

-let T:V-W be a linear map
-S and S’ are bases of V
-R and R’ are bases of W
-the matrices A=A(R,S) and A’=A(R’,S’) of T w.r.t. S & R and S’ and R’ respectively are equivalent:
A’ = P^(-1) A Q

Question 29

Similarity Proposition

Answer 29

-suppose T:V->V be linear
-let S and S’ be be bases of V
-the matrices B=B(S,S) and B’=B(S’,S’) of T, with respect to S in source and target and S’ in source and target respectively, are similar:
B’ = P^(-1) B P

Question 30

Injective

Definition

Answer 30

• suppose that f:X->Y is a mapping between two sets

- f is said to be injective if ∀ x,x’∈X , f(x)=f(x’) => x=x’

Question 31

Surjective

Definition

Answer 31

• suppose that f:X->Y is a mapping between two sets

- f is said to be onto or surjective if ∀ y∈Y, ∃ x∈X such that f(x)=y

Question 32

Bijection

Definition

Answer 32

• suppose that f:X->Y is a mapping between two sets

- f is said to be a bijection if it is injective andd surjective

Question 33

Injective and Kernel

Proposition

Answer 33

• let T:V->W be linear

- T is injective if and only if ker T = { |0 }

Question 34

Surjective and Image

Proposition

Answer 34

• let T:V->W be linear

- T is surjective if and only if imT = W

Question 35

Rank Nullity, Injective and Surjective

Propositions

Answer 35

a) T is injective if and only if n(T) = 0

b) T is surjective if and only if r(T) = dimW

Question 36

Row Rank and Column Rank

Answer 36

-we can define both a row rank and a column rank of a matrix A
row rank = dim (span(rows of A))
column rank = dim (span(columns of A))
-these are both equal to each other and the rank of A

Question 37

Rank of a Matrix

Theorem

Answer 37

-the row rank and column rank of a matrix both coincide with the number of non-zero rows of matrix B in echelon form obtained from A by row ops

Question 38

Rank-Nullity Formula and Matrices

Answer 38

• if A is an mxn matrix such that Ta(|v) = A |v
• the null space of A is the vector space ker(Ta)={|v∈F^n | A |v = 0 }
• the nullity of A, n(A)=dim(ker(Ta))
• the rank of A, r(A) = dim(im(Ta))

Question 39

How do you find the dimension of the null space of a matrix A?

Answer 39

• transform A into a matrix B in echelon form using row operations
• the dimension of the null space of A is equal to the number of columns in B with non-zero leading elements

Question 40

How to find the rank of a matrix A?

Answer 40

• reduce A to a matrix B in echelon form using row operations
• the rank of A is equal to the number of non-zero rows in B

Question 41

Canonical Form

Answer 41

-let T:V->W
-there are bases of V and W with respect to which the matrix of T has canonical form
-let {u1,…,un} be a basis of kerT
-then S={v1,…,vr;u1,…,un} is a basis of V
-and R={w1,…,wr;z1,..,zp} is a basis of W
-therefore with respect S and R, the matrix of T has the canonical form:
M = (Ir 0)
(0’ 0’’)
-where Ir is the rxr identity matrix whre r=rank(t) and 0,0’,0’’ are zero matrices of appropriate size

Question 42

How to find the Canonical Form

Answer 42

-to find the canonical form of a matrix of a linear transformation T:V->W, find bases with respect to which the matrix of T is canonical form:
1) Find a basis {u1, … , un} of ker T
2) extend to a basis S={v1,…,vr,u1,…,un} of V by adding vectors from the standard basis of V to the basis for kerT
3) transform this basis of V using T, T(ui)=0 since the set of ui vectors forms a basis of kerT and the vectors T(v1),…,T(vr) are automatically linearly independent since they form a basis of im(T), so we can extend them to a basis R={T(v1),..T(vr),z1,…,zp} of W
4) then w.r.t. S and R, the matrix M=M(R,S) of T will be in canonical form, M = P^(-1)AQ
where:
- A is the transformation matrix of T with respect to the standard bases of V and W
-P expresses the basis R in terms of the standard basis of W
-Q expresses the basis S in terms of the standard basis of V

Question 43

Linear Combinations of Linear Maps

Answer 43

-let V and W be F-vector spaces
-the set Hom(V,W) = { T:V->W | T is linear} of all linear maps from V to W is a vector space with the following operations:
i) if S,T : V->W are linear maps then S+T is such that
(S+T)(|v) = S(|v) + T(|v) for all |v∈V
ii) if T:V->W is a linear map and a∈F then aT is the mapping tom V to W defined by:
(aT) (|v) = a * T(|v)
-to prove show closure, i.e that:
(T+S)(|v+|w) = (T+S)(|v) + (T+S)(|w)
(T+S)(a|v) = a (T+S)(|v)