Survey
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
Download DEFINITION: A vector space is a nonempty set V of objects, called
Document related concepts
no text concepts found
Transcript
DEFINITION:
A vector space is a nonempty set V of
objects, called vectors, on which are defined two operations, called addition and
multiplication by scalars (real numbers),
subject to the following 10 axioms (or
rules):
1. The sum of ū and v̄, denoted by
ū + v̄, is in V.
2. ū + v̄ = v̄ + ū.
3. (ū + v̄) + w̄ = ū + (v̄ + w̄).
4. There is a zero vector 0̄ in V such
that ū + 0̄ = ū.
5. For each ū in V, there is a vector
−ū in V such that ū + (−ū) = 0̄.
6. The scalar multiple of ū by c, denoted by cū, is in V.
7. c(ū + v̄) = cū + cv̄.
8. (c + d)ū = cū + dū.
9. c(dū) = (cd)ū.
10. 1 · ū = ū.
These axioms must hold for all vectors
ū, v̄, and w̄ in V and all scalars c and d.
EXAMPLE:
x1
x2
: x1, . . . , xn ∈ R
1. Rn =
.
.
xn
2. The set Pn of polynomials of degree
at most n:
p̄(t) = antn + . . . + a2t2 + a1t + a0
where the coefficients an, . . . , a0 and the
variable t are real numbers.
3. The set of all real-valued functions
defined on R.
DEFINITION:
A subspace of a vector space V is a subset H of V that has 3 properties:
1. The zero vector of V is in H.
2. H is closed under vector addition.
That is, for each ū and v̄ in H, the sum
ū + v̄ is in H.
3. H is closed under multiplication by
scalars. That is, for each ū in H and
each scalar c, the vector cū is in H.
EXAMPLE:
The set consisting of only the zero vector 0̄ in a vector space V is a subspace of
V, called the zero subspace and written
as {0̄}.
WARNING:
R2 is not a subspace of R3, because R2
is not a subset of R3.
EXAMPLE:
The set
s
H = t : s and t are real numbers
0
is a subspace of R3.
THEOREM:
If v̄1, . . . , v̄p are in a vector space V, then
Span {v̄1, . . . , v̄p} is a subspace of V.
EXAMPLE:
Let
1
1
v̄1 = 2 v̄2 = 1 .
3
2
By the Theorem above
Span{v̄1, v̄2}
is a subspace of R3.
EXAMPLE:
Let H be the set of all vectors of the
form
4a − b
2b
a − 2b
a−b
where a and b are arbitrary scalars. Show
that H is a subspace of R4.
SOLUTION:
We have
4a − b
4
−1
2b
0
2
= a +b
a − 2b
1
−2
a−b
1
−1
| {z }
| {z }
v̄1
v̄2
We see that
H = Span{v̄1, v̄2}
therefore H is a subspace of R4 by the
Theorem above.
EXAMPLE:
Let H be the set of all vectors of the
form
a−b
b−c
c−a
b
where a, b and c are arbitrary scalars.
Find a set S of vectors that spans H or
show that H is not a vector space.
SOLUTION:
We have
a−b
1
−1
0
b−c
0
1
−1
= a
+b
+c
c−a
−1
0
1
b
0
1
0
| {z }
| {z }
| {z }
v̄1
v̄2
v̄3
and we see that H is a vector space and
{v̄1, v̄2, v̄3}
spans H.
EXAMPLE:
Let H be the set of all vectors of the
form
3a + b
4
a − 5b
where a and b are arbitrary scalars. Show
that H is not a vector space.
SOLUTION:
H is not a vector space, since 0̄ 6∈ H
(the second entry is always nonzero).
EXAMPLE:
Let H be the set of
form:
a
(a) b
0
a
(b) b , where
c
a
(c) a
1
a
(d) b , where
c
a
(e) b , where
c
all vectors of the
a≥0
a + 2b − c = 0
2a − b + c = 1.
Is H a subspace of R3?
SOLUTION:
(a) We have
1
0
a
b = a0 + b1,
0
0
0
which is a linear combination and therefore a subspace of R3.
(b) It is not a subspace, since it is not
closed under multiplication by a scalar.
In fact,
a
−a
(−1) b = −b ,
c
−c
and the first entry is negative, which
contradicts the initial assumption.
(c) It is not a subspace, since there is no
a zero vector (the last entry is always
nonzero).
(d) We have
0
1
a
a
b = b = a0 + b1,
2
1
a + 2b
c
which is a linear combination and therefore a subspace of R3.
(e) We have
a
a
,
b =
b
1 − 2a + b
c
which is not a subspace, since there is
no a zero vector. In fact,
0
a
= 0
b
0
1 − 2a + b
if and only if a = 0 and b = 0. But in
this case the last entry in nonzero, since
1 − 2a + b = 1 − 2 · 0 + 0 = 1.
Contradiction.
DEFINITION:
The null space of an m × n matrix A,
written as Nul A, is the set of all solutions to the homogeneous equation
Ax̄ = 0̄.
DEFINITION0:
The null space of an m × n matrix A is
the set of all x̄ in Rn that are mapped
into the zero vector 0̄ in Rm by the linear transformation
x̄ 7−→ Ax̄.
EXAMPLE:
Let
1 −2 −1
.
2 −3 −4
5
Determine if ū = 2 belongs to the
1
null space of A.
A=
SOLUTION:
Since
5
1 −2 −1
0
2 =
Aū =
,
0
2 −3 −4
1
ū is in Nul A.
EXAMPLE:
Let
1 −3 −2
.
−5 9 1
5
Determine if ū = 3 belongs to the
−2
null space of A.
A=
SOLUTION:
Since
Aū =
5
1 −3 −2
0
3 =
,
0
−5 9 1
−2
ū is in Nul A.
THEOREM:
The null space of an m × n matrix A
is a subspace of Rn. Equivalently, the
set of all solutions to a system Ax̄ = 0̄
of m homogeneous linear equations in n
unknowns is a subspace of Rn.
EXAMPLE:
Find a spanning set for the null space of
the matrix
−3 6 −1 1 −7
A = 1 −2 2 3 −1 .
2 −4 5 8 −4
SOLUTION:
We find the general solution of Ax̄ = 0̄:
1 −2 0 −1 3 0
[A 0̄] ∼ 0 0 1 2 −2 0 ,
0 0 0 0 0 0
therefore
x1 − 2x2 − x4 + 3x5 = 0
x3 + 2x4 − 2x5 = 0,
x1
2x2 + x4 − 3x5
x2
x
2
so
x3 = −2x4 + 2x5
x4
x4
x5
x5
1
−3
2
1
0
0
= x2 0 +x4 −2 +x5 2
,
0
1
0
0
0
1
| {z }
| {z }
| {z }
ū
v̄
so Nul A =Span {ū, v̄, w̄}.
w̄
DEFINITION:
The column space of an m×n matrix A,
written as Col A, is the set of all linear
combinations of the columns of A.
REMARK:
So, if A = [ā1 . . . ān], then
Col A = Span{ā1, . . . , ān}.
THEOREM:
The column space of an m × n matrix is
a subspace of Rm.
EXAMPLE:
Let
2 4 −2
A = −2 −5 7
3 7 −8
Find a nonzero vector in
nonzero vector in Nul A.
1
3.
6
Col A and a
SOLUTION:
1. Any column of A is anonzero
vector
2
in Col A. For example, −2 =
3
1
4
−2
2
= 1 −2 +0 −5 +0 7 +0 3 .
6
8
7
3
2. To find a nonzero vector in Nul
A, we row reduce the augmented matrix [A 0̄]:
1 0 9 0 0
[A 0̄] ∼ 0 1 −5 0 0 ,
0 0 0 1 0
therefore any vector
x1
x2
=
x̄ =
x3
x4
−9x3
5x3
x3
0
is in Nul A. For example, if we put x3 =
1, we get
−9
5
ū =
1
0
is in Nul A.
Related documents