4) Linear Maps And Vector Subspaces Flashcards
Matrix multiplication
L_A( r_1,…,r_n)
n n
( Σ a_1j •r_j,…, Σ a_mj•r_j)
j=1 j=1
WHEN IS A MAP
A LINEAR MAP
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
Proposition: linear map and matrix
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
Proposition of composition of linear map
If A,B are matrices and the product AB is defined, L_AB = L_A • L_B
Composition
Directional derivative and linear maps
And dot products?
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
Def of grad
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
Definition: vector subspace
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
Subspaces arise
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
Given an equation for a plane: finding subspace representing
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}
For a subspace with zero vector
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
Unique way of writing a vector in a subspace with vectors r_1,…,r_n
When linearly independent
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
Definition: spanning set for V
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
Theorem: for V a vector subspace
Spanning set , basis
Theorem for dims
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
FINDING A BASIS
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’
Finding a basis for a line
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)
Finding a basis for a plane
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
Proposition: For V and V’ subspaces of R
^p
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
Definition: for set of linearly independent vectors
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
Definition of the span of vectors
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.
Proposition: linearly independent vectors and showing a way to obtain intrinsic equation for subspace of R^p
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.
Proposition:
Image of a linear map
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)
Proposition:nullspace or kernel of a linear map
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.
Definition: of a linear map rank and null
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
The rank nullity theorem
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.
