Chapter 4 - Pinear Maps And Vector Subspaces Flashcards
Conditions for a map to be lineat
For a r, r’ in R^P and t in R,
L (r + r’) = L (r) + L (r’) and
L (tr) = tL (r)
Conditions of a vector subspace
0 in V
If r is in V and r’ is in V, then r + r’ is in V
If r is in V and t is a real number then tr is in V
Spanning set for V (a subspace of R^P)
A set {r1, …, rn} of elements of V spans V or is a spanning set for V if
every element can be written as
v = t1r1 + … + tnrn for some t1, … , tn in the real numbers
i.e. they can be written as a linear combination of the set of elements
When is r1, …, rn a basis for V?
If it spans V and is linearly independent
If V is a subset of V ‘, then
If dimV = dimV’, then
DimV < or equal to dimV’
V = V’
Dimension of V
The number of elements in a basis of V.
Every basis for V has the same number of elements.
Image of a linear map
Im (L) = {L (r) | r is in R^p}
This is a subspace of R^q
The elements of the image are all the linear combibations of the columns of A
Nullspace or kernel of a linear map
Ker (L) = {r in R^p | L (r) = 0}
This is a subspace of R^p
Rank of a linear map
Rank (L) = dim (im (L))
This is equal to the number of pivots
Nullity of a linear map
Null (L) = dim (ker (L))
The number of basic solutions in terms of which every solition can be expressed, uniquely, as a linear combination
Rank-Nullity Theorem
Let L: R^p to R^q
Rank (L) + null (L) = p
Let L be a linear map with nullity 0 and suppose that r1, r2 in R^p have L (r1) = L (r2). Then
r1 = r2
If A and B are matrices and the product AB is defined, then Lab =
Lab = La ° Lb = La (Lb (r))
