Download General Vector Spaces - Computer Sciences - Lecture Slides and more Slides Operating Systems in PDF only on Docsity!
GENERAL VECTOR SPACES AND SUBSPACES [4.1]
General vector spaces
ä So far we have seen special spaces of vectors of n dimensions – denoted by Rn. ä It is possible to define more general vector spaces A vector space V over R is a nonempty set with two operations:
- Addition denoted by ′+′. For two vectors x and y, x + y is a member of V
- Multiplication by a scalar For α ∈ R and x ∈ V , αx is a member of V. ä In addition for V to be a vector space the following 8 axioms must be satisfied [note: order is different in text]
K-
- Addition is commutative u + v = v + u
- Addition is associative u + (v + w) = (u + v) + w
- ∃ zero vector denoted bv 0 such that ∀u, 0 + u = u
- Any u has an opposite −u such that u + (−u) = 0
- 1 u = u for any u
- (αβ)u = α(βu)
- (α + β)u = αu + βu
- α(u + v) = αu + αv
- Show that the zero vector in Axiom 3 is unique, and the vector −u, (‘negative of u’), in Axiom 4 is unique for each u in V.
ä For each u in V and scalar α, 0 u = 0 α0 = 0 −u = (−1)u
Example: Let V be the set of all arrows (directed line segments) in three-dimensional space, with two arrows re- garded as equal if they have the same length and point in the same direction. Define addition by the parallelogram rule, and for each v in V , define cv to be the arrow whose length is c times the length of v, pointing in the same direction as v if c > 0 and otherwise pointing in the opposite direction.
- Show that V is a vector space
Solution: The definition of V is geometric, using con- cepts of length and direction. No xyz-coordinate system is involved. An arrow of zero length is a single point and represents the zero vector. The negative of v is (−1)v.
Need to verify all axioms. See text
K-
More examples
ä Set of vectors in R^4 with second component equal to zero. ä Set of all poynomials of degree ≤ 3 ä Set of all m × n matrices ä Set of all upper triangular matrices
K-
Subspaces
ä A subset H of vectors of V is a subspace if it is a vector space by itself. Formal definition:
ä A subset H of vectors of V is a subspace if
- H is closed for the addition, which means: x + y ∈ H for any x ∈ H, y ∈ H
- H is closed for the scalar multiplication, which means:
αx ∈ H for any α ∈ R, x ∈ H
Observation If H is subspace then:
- 0 belongs to H
- For any x ∈ H, the vector −x belongs to H
ä Every subspace is a vector space. [why?] ä Conversely, every vector space is a subspace (of itself and possibly of other larger spaces). ä The set consisting of only the zero vector of V is a subspace of V , called the zero subspace. Notation: { 0 }. Example: Polynomials of the form p(t) = α 2 t^2 + α 3 t^3 form a subspace of the space of polynomials of degree ≤ 3 Example: Triangular matrices ä Recall that the term linear combination refers to a sum of scalar multiples of vectors, and span{v 1 , ..., vp} denotes the set of all vectors that can be written as linear combinations of v 1 , · · · , vp.
If u and v represent any two vectors in Nul(A). Then
Au = 0 and Av = 0
ä To show that u + v is in Nul(A), we must show that A(u + v) = 0. Using a property of matrix multiplication, compute
A(u + v) = Au + Av = 0 + 0 = 0
ä Thus u + v is in Nul(A), and Nul(A) is closed under vector addition.
ä Finally, if α is any scalar, then
A(αu) = α(Au) = α(0) = 0
which shows that αu is in Nul(A). Thus Nul(A) is a subspace of Rn.
K-
ä There is no obvious relation between vectors in Nul(A) and the entries in A. ä We say that Nul(A) is defined implicitly, because it is defined by a condition that must be checked. ä No explicit list or description of the elements in Nul(A) is available, so... ä ... we to solve the equation Ax = 0 to produce an explicit description of Nul(A). Example: Find a spanning set for the null space of the matrix
A =
K-
Solution: first step is to find the general solution of Ax = 0 in terms of free variables. We know how to do this.
ä Get reduced echelon form of augmented matrix [A 0]:
x 1 − 2 x 2 − x 4 +3x 5 = 0 x 3 + 2x 4 − 2 x 5 = 0 0 = 0
ä x 2 , x 4 , x 5 are free variables, x 1 , x 3 basic variables.
ä For any selection of the free variables, can find a vector in Nul(A) by computing x 1 , x 3 in terms of these variables:
x 1 = 2x 2 + x 4 − 3 x 5 x 3 = − 2 x 4 + 2x 5
ä OK - but how can we write these using spanning vectors (i.e. as linear combinations of specific vectors?)
ä Solution - write x as:
x 1 2 x 2 +x 4 +3x 5 x 2 x 2 x 3 = − 2 x 4 +x 5 x 4 x 4 x 5 x 5
= x 2
u
+x 4
v
+x 5
w ä General solution is of the form x 2 u + x 4 v + x 5 w. ä Every linear combination of u, v, and w is an element of Nul(A). Thus {u, v, w} is a spanning set for Nul(A).
- Obtain the vector x of Nul(A)corresponding to the choice: x 2 = 1, x 4 = − 2 , x 5 = − 1. Verify that indeed it is in the null space, i.e., that Ax = 0
- For same example, find a vector in Nul(A)whose last two components are zero and whose first component is 1. How many such vectors are there (zero, one, or inifintely many?)
Notes:
ä 1. The spanning set produced by the method in the example is automatically linearly independent
- Show this [write that the lin. combination is zero. & look at top of previous page]
ä 2. When Nul(A)contains nonzero vectors, the number of vectors in the spanning set for Nul(A) equals the number of free variables in the equation Ax = 0.
K-
Column Space of a matrix
Definition: The column space of an m × n matrix A, written as Col(A) (or C(A)), is the set of all linear com- binations of the columns of A. If A = [a 1 · · · an], then
Col(A) = span{a 1 , ..., an}
Theorem: The column space of an m × n matrix A is a subspace of Rm. ä A typical vector in Col(A) can be written as Ax for some x [the notation Ax stands for a linear combination of the columns of A]. That is, Col(A) = {b : b = Ax for some x in Rn}
K-
ä The notation Ax for vectors in Col(A) also shows that Col(A) is the range of the linear transformation x → Ax.
ä The column space of an m × n matrix A is all of Rm^ if and only if the equation Ax = b has a solution for each b in Rm
A =
,^ u^ =
, v =
a. Determine if u is in Nul(A). Could u be in Col(A)?
b. Determine if v is in Col(A). Could v be in Nul(A)?
ä General remarks and hints:
- Col(A) is a subspace of Rm^ [m = 3 in above example]
- Nul(A) is a subspace of Rn^ [n = 4 in above example]
- To verify that a given vector x belongs to Nul(A) all you need to do is check if Ax = 0
- To verify if b ∈ Col(A) all you need to do is check if the linear system Ax = b has a solution.
