4) Linear Maps And Vector Subspaces

Question 1

Matrix multiplication

Answer 1

L_A( r_1,…,r_n)
n n
( Σ a_1j •r_j,…, Σ a_mj•r_j)
j=1 j=1

Question 2

WHEN IS A MAP

A LINEAR MAP

Answer 2

A map L: R^p -> R^q is linear if
For all vectors r and r’ ∈ R^p and t∈R:

1) L(r + r’) = L(r) + L(r’)
2) L(tr) = tL(r)

Every linear map by matrix, bijective
q * p

Question 3

Proposition: linear map and matrix

Answer 3

If A LINEAR map L: R^p -> R^q

THEN equal to a unique q x p matrix

Standard basis -> columns of linear map
Proof:
Any vector expressed in R^p in terms of a standard basis, since linear map properties of addition of L(e_i), which each in R^q, written as standard basis for R^q regrouped shows matrix multiplication

Question 4

Proposition of composition of linear map

Answer 4

If A,B are matrices and the product AB is defined, L_AB = L_A • L_B

Composition

Question 5

Directional derivative and linear maps

And dot products?

Answer 5

Directional derivative of F at (u,v) in the direction (a,b)
For x=F(u,v) a smooth real valued function, D(F)(u,v) = [partial derive x wrt u, partial deriv x wrt v] row matrix

So it’s a linear map as represented by matrix
D(F)(u,v) = dot product of (a,b) • gradient =
a (partial F wrt u) + b (partial F wrt v)

Directional deriv dot product of vector with grad f

Generally unit vector
E.g. Cos theta sin theta

Question 6

Def of grad

Answer 6

Gradient of F ,
Nabla F = grad F
= ( ∂F/ ∂u_1,…, ∂F/∂u_p) ∈ R^p

Where x= F(u_1,…, u_p) is a smooth function of p variables

Question 7

Definition: vector subspace

Answer 7

Let V be a subset of R^p. Then V us a vector subspace of R^p if:
1) (0 vector) ∈V (always through 0)

2) if vectors r,r’∈V then vector r+r’ ∈V
3) if vector r ∈V and t ∈R, vector tr∈V

Question 8

Subspaces arise

Answer 8

Trivial subsets {0vector} and R^p are Subspaces of R

R^2: straight lines through origin
ax+by =0 where a^2 + b^2 not equal to 0,
V ={tvector(r) : t∈R}

R^3: straight lines and planes through origin
Planes can be written ax+by+cx=0
Where a^2+ b^2+c^2 not equal to
As linear combo of linearly independent
V= {sr+tr’ : s,t∈R}
Indep i.e. sr+tr’=vector(0) only for (0,0)

In R^3 lines through the origin can be written as {tr : t∈R} for non zero vector r and described by two linear equations ax+by+cx=0 and a’x+b’y+c’z=0 which don’t coincide and intersection -» line

Question 9

Given an equation for a plane: finding subspace representing

Answer 9

Eg find two linearly independent vectors in the plane by inspection ( cross product doesn’t equal 0)
V as linear combo
Or a line is:
V ={tr : t∈R}

Question 10

For a subspace with zero vector

Answer 10

V of R^p. With r_1,…r_n ∈ V,
Known such that V= linear combo of those n vectors

If 0 vector ∈ {r_1,…,r_n} are linearly dependent!!!

Two non-zero vectors r,r’ in R^p are linearly dependent if and only if each a scalar multiple of each other

Question 11

Unique way of writing a vector in a subspace with vectors r_1,…,r_n
When linearly independent

Answer 11

Proposition: let r_1,…,r_n ∈ R^p be linearly independent. Suppose that t_1,…,t_n & t’_1,…,t’_n are scalars such that
t_1r_1 + …+t_nr_n = t’_1r_1 + …+t’_nr_n
Implies t_1=t’_1,… t_n=t’_n

Proof:
If linearly independent then only equal zero when (t_1 -t’_1) =0 i.e. Equal

Question 12

Definition: spanning set for V

Answer 12

For V subspace of R^p. Then a subset {r_1,…,r_n} of elements of V. Spans V (a spanning set) if every element vector(v) in V can be linear combo
Vector(v) =t_1r_1 + …+t_nr_n for t_1,…,t_n in R

r_1,…,r_n form a basis for V if spanning set spans V and they are linearly independent

Question 13

Theorem: for V a vector subspace
Spanning set , basis
Theorem for dims

Answer 13

Theorem: Let V be a vector subspace of R^p:
1) given linearly independent vectors r_1,…,r_k in V there are FINITELY MANY FURTHER r_(k+1),…,r_n st they form a basis

2) Given a spanning set r_1,…,r_m there is a finite subset r_1,…,r_p that is a basis for V
3) Every basis for V has same dimension and elements

Dim of R^p is p
dim of {0 vector} is 0 and basis is the empty set

Standard basis spans R^p and is linearly independent

Question 14

FINDING A BASIS

Answer 14

R^2: (a,b) not equal to 0, on its own linearly independent to (-b,a)
For example: (a,b) (-b,a) basis

R^3: suppose given 2 check linearly independent by cross then basis is r,r’, r x r’

Question 15

Finding a basis for a line

Answer 15

For au+be=0 a line a^2 + b^2 not equal to 0 be a line through the origin in R^2

V ={(u,v) : au+bv=0}
Clearly (-b,a) in V
For any (u,v) in V
(u,v)• (a,b) =0
Perpendicular so scalar multiple of (-b,a) i.e. (u,v) =t(-b,a) for t in R
(-b,a) spans V and it’s a basis
As a line au+bv=0 is a line in direction (-b,a)

Question 16

Finding a basis for a plane

Answer 16

For a plane au+bv+cw=0 where a^2 + b^2+ c^2 not equal to 0.
Find a basis for V={(u,v,w) :au+bv+cw =0 } subset of R^3

Find two vectors in the plane
Linearly independent if cross product not equal 0 vector

Try third and then determinant of 3 vectors not equal to 0 then linearly independent

If we are given basis we can find plane equation by cross product of two vectors is a normal

Question 17

Proposition: For V and V’ subspaces of R

^p

Answer 17

Let V and V’ be subspaces if R^p:

If V is a subset of V’

Then dim V is less than or equal to dim V’

If dim V = dim V’ then V = V’

Proof: we can express a basis b_1,…,b_n as a basis for V and as is a subset for V’ and they’d elinearly independent they can be extended to a basis b_1,…,b_n,b_n+1,…,b_n’ by a the tiren. So we have dimensions bigger or equal to if we have equal dimensions it means it’s a basis also for V’ so spans V’ and equal

Question 18

Definition: for set of linearly independent vectors

Answer 18

Let r_1,…,r_n be vectors in R^p. The set {r_1,..,r_n} is linearly independent if the only scalars t_1,..,t_n such that t_1+… +t_n r_n = vector(0)
Are t_1=0,…,t_n=0

If one of the vectors is vector(0) then we know it’s linearly dependent

Question 19

Definition of the span of vectors

Answer 19

Let r_1,..,r_k be vectors in R^p. Then the span of r_1,…,r_k is the set of all linear combinations of the r_1,..,r_k. That is span { r_1,..,r_k} = {t_1r_1 +…+ t_kr_k | t_1,…,t_k in R}

THE SPAN IS ALWAYS A SUBSPACE

Subspaace of R^p or has dimensión less than or equal to

As. If not linearly independent then can span by less vectors.

Question 20

Proposition: linearly independent vectors and showing a way to obtain intrinsic equation for subspace of R^p

Answer 20

Let r_1,…,r_p-1 be linearly independent vectors in R^p. Write V for the subspace spanned by r_1,r_2,…,r_p-1. Then V is defined by the equation:

Det ( [row 1: r_1..]
[Row 2: r_2]
[....]
[Row p-1: r_p-1
[Row p: x_1 x_2 ... x_p] )
=0
VECTORS AS ROWS

I.e. det is 0 iff the rows are linearly dependent since the first p-1 are the rest are only if vector x is. That’s equivalent to the final row
being in the span of the first p-1 rows.

This gives the intrinsic equation of a plane through 0 by finding the determinant.

Question 21

Proposition:

Image of a linear map

Answer 21

L:R^p -> R^q be a linear map. Then the image of L, defined by im(L) = { L(vector(r)) | vector r in R^p}
Is a subspace of R^q.

Proof:
We show by subspace criterion that 0_q is in im(L), for r’_1and r’_2 in im(L) Given by..
We have that r’_1 + r’_2 = L(r_1 + r_2) is in (L) and finally for scalar t tr’_1 = tL(r_1) = L(tr_1) in im(L)

Question 22

Proposition:nullspace or kernel of a linear map

Answer 22

Let L: R^p -> R^q be a linear map. Then the nullspace or KERNEL of L defined by Ker(L)
= {vector(r) in R^p| L(vector(r)) = vector(0)}

Is a SUBSPACE of R^p
P not q

Proof: subspace criterion, vector 0_p is in nullspace, for two vectors in Ker(L) their sum is also: L(r_1 + r_2) = L(r_1) + L(r_2) = 0_q + 0_q and similarly for scalar of vector in kernel.

Question 23

Definition: of a linear map rank and null

Answer 23

Let L:R^p -> R^q be a linear map

Rank(L) is the dimension of the image of L
Null(L) is the dimension of the nullspace or kernel

Null(L) is an integer and ker(L) the nullspace is a vector subspace

Question 24

The rank nullity theorem

Answer 24

Let L: R^p -> R^q be a linear map. Then rank(L) + null(L) =p

Proof:
Basis for Ker(L) can be shown to extended for a basis of R^p…
We claim and show that L(b_n+1),..L(b_p) , the image of these vectors, is a basis for the image im(L). By criterion of linearly independent and showing spanning. Thus dimensions must give p.

Question 25

Corollary of the rank nullity theorem

Answer 25

For a linear map L:R^p -> R^q

Rank(L) is LESS THAN OR EQUAL TO min(p,q)

E.g. If a linear map has rank 0 then it’s image is the zero subspace if R^q and it’s basis is the empty set. So L(r) =0 for all vectors r in R^p and it’s the zero linear map.

Question 26

Sets of solutions/ solution sets for NULLSPACE

Answer 26

NULLSPACE is the set of solutions to the system of linear equations
Set of vectors x
Vector(x) = (x_1,..x_p)

a_11 x_1 + a_12 x_2 +…+ a_1p x_p =0
a_21 x_1 + a_22 x_2+…+ a_2p x_p =0
….
aq1 x_1 + aq2 x_2 +…. + a_qp x_p =0

Row reductions on these k equations in p unknowns is possible for k less than or equal to q.
When we have k is the number of pivots we have p-k free variables and p-k basic sols

The nullity is the number of basic solutions for which every solution can be expressed uniquely as

Question 27

Sets of solutions/ solution sets for The IMAGE

Answer 27

The image of map L:R^p -> R^q is the set of (y_1,…,y_q) such that

a_11 x_1 + a_12 x_2 +…+ a_1p x_p =y_1
a_21 x_1 + a_22 x_2+…+ a_2p x_p =y_2
….
a_q1 x_1 + aq2 x_2 +…. + a_qp x_p =y_q

Has a solution (x_1,..,x_p) in R^p

The rank is the dimension of the SUBSPACE OF R^q SPANNED BY THE COLUMNS OF A

no row reduction only find linearly independent columns

The image set of vectors x in R^p is written as a linear combo of this

Same nullspace after reduction not same column span

Question 28

Writing a general vector in its basis form

Answer 28

A general vector x in R^p can be written as x_1 e_1 +…+ x_p e_p where e_i are standard basis

Useful for matrix multiplication

Question 29

Proposition: linear map with nullity zero with L(r_1) =L(r_2)

Answer 29

Proposition: let L: R^p -> R^q be a linear map with nullity map with nullity zero and suppose that r_1,r_2 in R^p have L(r_1) = L(r_2). Then r_1 = r_2.

IF THE NULLITY IF L IS 0 then L IS INJECTIVE
Proof: shown as by linearity we have L(r_1) -L(r_2) =0 so r_1 - r_2 is in ker(L) and therefore as the nullity is 0 vectors taken away equal 0 and thus equal.

E.g. If A has nullity zero only solution to equations is a unique solution

Question 30

Example given a map: find nullity and rank

Answer 30

Check that it’s a linear map.

Find the rank by linearly independent vectors and dimension of column space.

Find the nullity from reduction.

Check by RNT.

Question 31

Finding inverse matrices

Answer 31

•row operations
Augmented matrix perform row operations
•minors and cofactors
Minor: determinant if I,jth
Cofactor: the determinant multiplied by (-1)^(I+j)
Then take the matrix of cofactors: TAKING TRANSPOSE AND DIVIDING BY DETERMINANT

Question 32

Example of the determinant matrix of polar coordinates finding the rank and nullity

Answer 32

To find the nullity we can perform row reduction: by adding multiples of costheta or sintheta in one case we here each not equal to 0.
•Consider both not equal to 0. That is by multiplying through rows by them and performing row reduction. Then dividingwe obtain the identity matrix. Hence nullity is 0
• if they are equal to 0 we consider each case separately: costheta not equal to 0 or sintheta not equal to 0. By this we have identity matrix again
•case r us equal to 0 rank is 1 and nullity q

Hence for all theta F(0,theta) = (0,0)
The nullity is zero and the rank is two at all points except the points (0,theta) at which points the nullity is one and the rank is 1.

Question 33

Example 4.37

Answer 33

Smooth map.

Question 34

Example 4.38

Answer 34

Look

Question 35

Example 4.4p

Answer 35

Look

Question 36

Lemma: for two linear maps such that their composition is the 0 map then we have..

Answer 36

Let L:R^p -> R^q and L’: R^q -> R^r be linear maps such that L’•L is the 0 map R^p -> R^r.

Then im(L) is a subset of ker(L’)

Proof: from the assumption we can show any element of image(L) is in kernel of L’

Question 37

Last example chapter 4

Answer 37

Look