Download Subspaces of Rn (2.8) 1. Definition Definition A subspaces of Rn is a

Subspaces of R n (2.8)
1. Definition
Definition A subspaces of R n is a set H in R n that has three properties:
a. The zero vector is in H.
b. For each u and v in H, the sum u v is in H.
c. For each u in H and each scalar c, the vector cu is also in H.
Note that Property b. and c. also mean a subspace is closed under vector addition and scalar
multiplication. Clearly, R n is also a subspace of R n .
Example Let v 1 and v 2 be in R n . Let H Span v 1 , v 2 . Show that H is a subspace of R n .
Check conditions a., b., and c.:
a. 0v 1 0v 2 0 is in H.
b. Let u )v 1 *v 2 and v c 1 v 1 c 2 v 2 be in H.
Thenu v Ÿ)v 1 *v 2 Ÿc 1 v 1 c 2 v 2 Ÿ) c 1 v 1 Ÿ* c 2 v 2 is in H.
c. Let u )v 1 *v 2 and c be a scalar. Then
cu cŸ)v 1 *v 2 c)v 1 c*v 2 is in H.
Therefore, H is a subspace of R n .
Note that H Span v 1 , v 2 , v 3, T , v l is a subspace of R n .
Example Let v 1 and v 2 be in R n . Let S subspace of R n .
u tv 1 v 2 , t is any real number
. Determine if S is a
Check conditions c.:
a. Let u )v 1 v 2 and c p 1. Then
cu cŸ)v 1 v 2 Ÿc) v 1 cv 2 p tv 1 v 2
Therefore, S is not a subspace of R n .
Example Page 173: 1,2
2. Column Space and Null Space of a Matrix
Definition The column space of a matrix A is the set ColŸA of all linear combinations of the columns of
A.
Definition The null space of a matrix A is the set NulŸA of all solutions to the homogeneous equation
Ax 0.
Note that if A is an m • n matrix then the column space ColŸA is a subspace of R m and the null space
NulŸA is a subspace of R n .
1
"3 "4
1
Example Let A a.
b.
c.
d.
"4 6
"2
"3 7
6
3
, b
5
, v
3
3
"4
"1
Determine if b is in the column space of A.
Determine if v is in the null space of A.
Find the column space of A.
Find the null space of A.
a. b is in the column space of A if and only if there exists x in R 3 such that Ax b. Check if Ax b is
consistent.
"3 "4 3
1
Ab
"4 6
"2 3
"3 7
6
1 "3
Gaussian Elimination
3
0 "6 "18 15
®
"4
"4
0
0
0
0
Since the linear system Ax b is consistent, b is in the column space of A.
b. v is in the null space of A if and only if Av 0. Check
"3 "4
1
Av 5
"4 6
"2
3
"3 7
6
"1
0
0
0
v is in the null space of A.
"3
1
c. ColŸA Span
"4
,
"4
,
6
"3
"2
7
6
d. Find all x such that Ax 0 :
1 "3
A0
®
"4
0
0 "6 "18 0
0
0
0
,
0
x 2 "3x 3
x 1 3Ÿ"3x 3 4x 3 "5x 3
"5t
, x
"3t
t
"5
t
"3
1
"5
NulŸA Span
"3
.
1
3. Basis for a Subspace
Definition A basis for a subspace H of R n is a linearly independent set in H that spans H.
Standard basis for R n :
e1, e2, T , en
where e i is the ith column of I n . Clearly, e 1 , e 2 , T , e n are
v1
linearly independent. For every vector v B
vn
2
in R n ,
v v 1 e 1 v 2 e 2 C v n e n .
Example Give a basis of R n (other than standard basis).
1
1
0
Consider n 4. Let v 1 1
, v2 0
1
, v3 0
0
0
1
1
0
1
, v4 1
1
.
1
Let A v 1 , v 2 , v 3 , v 4 . Clearly, Ax 0 has only trivial solution and for every b in R 4 Ax b has a
unique solution . Hence, v 1 , v 2 , v 3 , v 4 are linearly independent and Span v 1 , v 2 , v 3 , v 4 R 4 .
Therefore, v 1 , v 2 , v 3 , v 4 forms a basis for R 4 . Similarly, v 1 , v 2 , T v n forms a basis for R n
1
Example Let A "3 "4
"4 6
"2
"3 7
6
. Find a basis for the column space of A and a basis for the null
space of A.
"3
1
ColŸA Span
,
"4
6
"3
1 "3
Since A ®
ColŸA .
3
0
,
7
"4
0 "6 "18
0
"4
0
6
"3
1
,
.
"2
"4
"3
,
6
7
is linearly independent. So, it forms a basis for