Chapter 5 Chapter Content
Chapter 5 Chapter Content
Chapter 5 Chapter Content
2. Subspaces
3. Linear Independence
• The only requirement is that the ten vector space axioms be satisfied.
n
Example ( R Is a Vector Space)
The set V = R n with the standard operations of addition and scalar multiplication
is a vector space. (Axioms 1 and 6 follow from the definitions of the
n
standard operations on R ; the remaining axioms follow from Theorem 4.1.1.)
The three most important special cases of R n are R (the real numbers), R 2
(the vectors in the plane), and R 3 (the vectors in 3-space).
Example (2×2 Matrices)
Show that the set V of all 2×2 matrices with real entries is a vector space if
vector addition is defined to be matrix addition and vector scalar
multiplication is defined to be matrix scalar multiplication.
u11 u12 v11 v12
Solution: Let u and v v
u21 u22 21 v22
u11 v11 u12 v12
(1) we must show that u + v is a 2×2 matrix. uv
u v u
21 21 22 22 v
(2) Want to show that u + v = v + u
u11 v11 u12 v12
uv vu
u v u
21 21 22 22 v
(3) Similarly we can show that u + ( v + w ) = ( u + v )+ w.
Thus, the set V of all 2×2 matrices with real entries is a vector space.
Example: Given the set of all triples of real numbers ( x, y, z ) with the operations
( x, y, z ) ( x ', y ', z ') ( x x ', y y ', z z ') and
k ( x, y, z ) (kx, y, z )
(1) If (x, y, z) and (x’, y’, z’) are triples of real numbers, so is
(x, y, z) + (x’, y’, z’) = (x + x’, y +y’, z + z’).
(2) (x, y, z) + (x’, y’, z’) = (x + x’, y + y’, z + z’)= (x’, y’, z’) + (x, y, z).
(3) (x, y, z) + [(x’, y’, z’) + (x’’, y’’, z’’)] = (x, y, z) + [(x’, y’, z’) + (x’’, y’’, z’’)].
(5) For each positive real x, (-x, -y, -z) acts as the negative:
(x, y, z) + (-x, -y, -z) = (-x, -y, -z) + (x, y, z) =(x, y, z)
(6) If k is a real and (x, y, z) is a triple of real numbers, then k (x, y, z) = (kx,
y, z) is again a triple of real numbers.
(7) k[(x, y, z) + (x’, y’, z’)] = (k(x+x’), y+y’, z+z’) = k(x, y, z) + k(x’, y’, z’)
Thus, the set of all triples of real numbers ( x, y, z ) with the operations is a
vector space under the given operation.
Example (Not a Vector Space)
k u = (k u1, 0)
There are values of u for which Axiom 10 fails to hold. For example, if u = (u1,
u2) is such that u2 ≠ 0,then
It’s easy to check that all the vector space axioms are satisfied.
(a) 0 u = 0
(b) k 0 = 0
(c) (-1) u = -u
(d) If k u = 0 , then k = 0 or u = 0.
5.2 Subspaces
Definition
A subset W of a vector space V is called a subspace of V if W is itself a
vector space under the addition and scalar multiplication defined on V.
Theorem 5.2.1
Remark
Theorem 5.2.1 states that W is a subspace of V if and only if W is a closed under
addition (condition (a)) and closed under scalar multiplication (condition (b)).
Example
All vectors of the form (a, 0, 0) is a subspace of R3.
(a, 0, 0) + (b, 0, 0) = (a + b, 0, 0)
k(a, 0, 0) = (ka, 0, 0)
Let W be the set of all points (x, y) in R2 such that x ≥ 0 and y ≥ 0. These are the
points in the first quadrant.
The set W is not a subspace of R2 since it is not closed under scalar multiplication.
For example, v = (1, 1) lines in W, but its negative (-1)v = -v = (-1, -1) does not.
Subspaces of Mnn
The set of n×n diagonal matrices forms subspaces of Mnn, since each of these
sets is closed under addition and scalar multiplication.
The set of n×n matrices with integer entries is NOT a subspace of the vector space
Mnn of n×n matrices.
This set is closed under vector addition since the sum of two integers is again an
integer. However, it is not closed under scalar multiplication since the product ku
where k is real and a is an integer need not be an integer. Thus, the set is not a
subspace.
Solution Space
Theorem 5.2.2
If Ax = 0 is a homogeneous linear system of m equations in n unknowns, then
the set of solution vectors is a subspace of Rn.
Definition
A vector w is a linear combination of the vectors v1, v2,…, vr if it can be
expressed in the form
w = k1v1 + k2v2 + · · · + kr vr
Since
Solution.
In order for w to be a linear combination of u and v, there must be scalars k1
and k2 such that w = k1u + k2v;
k1 + 6k2 = 9
2k1+ 4k2 = 2
-k1 + 2k2 = 7
w = -3u + 2v
Similarly, for w‘ to be a linear combination of u and v, there must be scalars k1
and k2 such that w'= k1u + k2v;
or
k1 + 6k2 = 4
2 k1+ 4k2 = -1
- k1 + 2k2 = 8
Theorem 5.2.3
If v1, v2, …, vr are vectors in a vector space V, then:
(b) W is the smallest subspace of V that contain v1, v2, …, vr in the sense that
every other subspace of V that contain v1, v2, …, vr must contain W.
Definition
If S = {v1, v2, …, vr} is a set of vectors in a vector space V, then the subspace
W of V containing of all linear combination of these vectors in S is called the
space spanned by v1, v2, …, vr, and we say that the vectors v1, v2, …, vr
span W. To indicate that W is the space spanned by the vectors in the set S
= {v1, v2, …, vr}, we write
Solution
Is it possible that an arbitrary vector b = (b1, b2, b3) in R3 can be expressed as
a linear combination b = k1v1 + k2v2 + k3v3 ?
Or
k1 + k2 + 2k3 = b1
k1 + k3 = b2
2k1 + k2 + 3 k3 = b3
This system is consistent for all values of b1, b2, and b3 if and only if the
coefficient matrix has a nonzero determinant.
However, det(A) = 0, so that v1, v2, and v3, do not span R3.
Theorem 5.2.4
If S = {v1, v2, …, vr} and S′ = {w1, w2, …, wr} are two sets of vector in a vector
space V, then