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Linear Algebra Theorem: Spanning Sets and Independent Sets, Study notes of Linear Algebra

Two important theorems in linear algebra: the first theorem states that a spanning set of vectors can be reduced to a basis, and the second theorem states that an independent set of vectors can be extended to a basis. The document also includes examples and proofs.

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2009/2010

Uploaded on 04/12/2010

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Math 311 Lecture 13
THEOREM FROM LAST TIME. If v1, v2, ..., vn span W, then
some subset of v1, v2, ..., vn is a basis for W. Moreover,
the basis subset can be chosen so that each deleted
vector vi depends on (is a linear combination of) earlier
basis vectors (i.e., basis vectors v1, v2, ..., vi-1).
C If the rref of the matrix [v1 | v2 | ... | v5 | 0 ] of the homogeneous
system is
x
1v1+x2v2+x3v3+x4v4+x5v5=0
v1v2v3v4v5
î
110200
001300
000010
x1=x22x4
x3=3x4
x5=0
Then the homogeneous system becomes:
(x22x4)v1+x2v2+(
3x4)v3+x4v4+0v5=0
Since x2 and x4 are arbitrary, v2 and v4 are dependent. To get v2, set
x2=1 and x4=0. This makes x3=x4=x5=0. Thus the system becomes
v1+v2=0 and v2=-v1. This process always writes a dependent vector
in terms of earlier vectors.
The previous theorem says that spanning sets can be
reduced to a basis by possibly deleting vectors. The
next theorem says that an independent set can be
increased to a basis by possibly adding vectors.
THEOREM. If v1, v2, ..., vk are independent vectors of a
space V, then v1, v2, ..., vk can be extended to a basis
by adding 0 or more vectors from the standard (or any
other) basis.
PROOF. Add the basis vectors to the end of the list to get
v1, v2, ..., vk, s1, s2, ..., sn. This set certainly spans V
since the standard basis vectors do. Applying the
above procedure then deletes dependent vectors to
form a basis. The deleted vectors are dependent on
earlier vectors. Since the vi's are independent, they are
not dependent on earlier vectors and so they are not
deleted. Hence the resulting basis consists of the v1, v2,
..., vk plus possibly some added vectors from the
standard basis. E
CWhich vectors of the standard basis should be added to
the set {v1, v2} = {[1,1,1], [2,1,1]} to form a basis?
Add s1 = [1,0,0], s2 = [0,1,0], s3 = [0,0,1], and convert to
columns. Reduce the augmented matrix for the
homogeneous system .
x
v1+yv2+us1+vs2+ws3=0
v1 v2 s1 s2 s3
é .
121000
110100
110010
1010 2 0
01 1 010
00 0 110
Answer: Add s2 which gives the basis {v1, v2, s2}.
COROLLARY. Suppose V is a vector space.
D If W is a maximal set of independent vectors (i.e., it
can't be extended to a larger set of independent
vectors) then W is a basis.
E If W is a minimal set of vectors which span V (i.e.,
no proper subset of W spans V) then W is a basis.
PROOF. D By the second theorem, W can be extended to a
basis. By hypothesis, no strictly larger set is
independent. Hence it must be a basis.
E By the first theorem, a subset of W is a basis. By
hypothesus, no strictly smaller set spans V. Hence W
is a basis. E
THEOREM. Suppose U and W are subsets of a vector space
V and suppose U has u elements and W has w
elements.
D If U is independent and W spans V, then u w.
E If U and W are both bases, then u = w.
PROOF OF (E FROM D. If U and W are both bases, then U
is independent and W spans V, hence by D u w.
Similarly w u and hence u = w.E
DEFINITION. The dimension n of a vector space V is the
number of elements in any basis of V. If V = {0}, the
dimension is 0.
COROLLARY. Suppose W is a subset of a vector space V,
W has w elements and V has dimension n.
D W independent implies w n; w > n implies W is
dependent.
E W spans V implies w ! n; w < n implies W doesn't span
V.
F W independent and w = n implies W is a basis.
G W spans V and w = n implies W is a basis.
PROOF. D and E follow the the theorem above. F and G
follow from the corollary above. E
C{t3, t2, t, 1} is a basis for P3, the vector space of
polynomials of degree 3. Thus P3 has dimension 4.
Explain why each the following cannot be a basis for
P3. {t3+1, t2+1, t+1} {t3+1, t31, t2, t, 1}
Page 102 Hw 9 Answers
4(2). (c+d)
u = c
u
d
u fails. 6(2). u
-u = 0 for some -u fails.
8(7). V = {r>0}, u
v = uv (ordinary multiplication), r
u = ur.
(1) Prove u
v = v
u. u
v = uv = vu = v
u
(2) u
(v
w) = u(v
w) =u(vw) = (uv)w = (u
v)w= (u
v)
w
(3) 0 =1. Prove u
0 = u: u
0 = u1 = u.
(4) -u = 1/u. Prove u
-u = 0: u
-u = u
(1/u) = 1 = 0
(6) (c+d)
u = u(c+d) = (uc)(ud) =(c
u)(d
u) = c
u
d
u
(7) c
(d
u) =(d
u)c = (ud)c =udc =(cd)
u. (8) 1
u = u1 =u
10(7). V = {0}. 0
0 = 0, r
0 = 0.
(1) u
v = 0
0 = 0 = 0
0 = v
u
(3) 0 = 0. Prove u
0 = u: u
0 = 0
0 = 0 = u
(4) -u = u. Prove u
-u = 0: u
-u = 0
0 = 0
(5) c
(u
v) = c
(0
0) = c
0 = 0 = 0
0 = c
u
c
v
(6) (c+d)
u = (c+d)
0 = 0 = 0
0 = c
ud
u
(7) c
(d
u) = c
0 = 0 = (cd)
u
(8)1
u = 0 = u
12(2). V = {x
L
R: x>0} u
v = uv
1, c
u = u.
The following fail: u
(v
w) = (u
v)
w, For some 0, u
0 = u ,
u
-u = 0 for some -u, (c+d)
u=c
u
d
u
14(2). V = R, u
v = 2u
v, c
u = cu.
The following fail: u
v = v
u, u
(v
w) = (u
v)
w
For some 0, u
0 = u, u
-u = 0 for some -u,
(c+d)
u=c
u
d
u

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Math 311 Lecture 13

THEOREM FROM LAST TIME. If v 1 , v 2 , ..., vn span W, then some subset of v 1 , v 2 , ..., vn is a basis for W. Moreover, the basis subset can be chosen so that each deleted vector vi depends on (is a linear combination of) earlier basis vectors (i.e., basis vectors v 1 , v 2 , ..., vi-1). C (^) If the rref of the matrix [v 1 | v 2 | ... | v 5 | 0 ] of the homogeneous system x 1 v 1 + x 2 v 2 + x 3 v 3 + x 4 v 4 + x 5 v 5 = 0 is v 1 v 2 v 3 v 4 v 5

Ó

  

1 − 1 0 2 0 0 0 0 1 − 3 0 0 0 0 0 0 1 0

  

x 1 = x 2 − 2 x 4 x 3 = 3 x 4 x 5 = 0 Then the homogeneous system becomes: ( x 2 − 2 x 4 ) v 1 + x 2 v 2 + ( 3 x 4 ) v 3 + x 4 v 4 + 0 v 5 = 0 Since x 2 and x 4 are arbitrary, v 2 and v 4 are dependent. To get v 2 , set x 2 =1 and x 4 =0. This makes x 3 = x 4 = x 5 =0. Thus the system becomes v 1 + v 2 =0 and v 2 =- v 1. This process always writes a dependent vector in terms of earlier vectors. The previous theorem says that spanning sets can be reduced to a basis by possibly deleting vectors. The next theorem says that an independent set can be increased to a basis by possibly adding vectors. THEOREM. If v 1 , v 2 , ..., vk are independent vectors of a space V, then v 1 , v 2 , ..., vk can be extended to a basis by adding 0 or more vectors from the standard (or any other) basis. P ROOF. Add the basis vectors to the end of the list to get v 1 , v 2 , ..., vk, s 1 , s 2 , ..., s (^) n. This set certainly spans V since the standard basis vectors do. Applying the above procedure then deletes dependent vectors to form a basis. The deleted vectors are dependent on earlier vectors. Since the v (^) i's are independent, they are not dependent on earlier vectors and so they are not deleted. Hence the resulting basis consists of the v 1 , v 2 , ..., vk plus possibly some added vectors from the standard basis. E CWhich vectors of the standard basis should be added to the set {v 1 , v 2 } = {[1,1,1], [2,1,1]} to form a basis? Add s 1 = [1,0,0], s 2 = [0,1,0], s 3 = [0,0,1], and convert to columns. Reduce the augmented matrix for the homogeneous system xv 1 + yv 2 + us 1 + vs 2 + ws 3 = 0. v 1 v 2 s 1 s 2 s 3

È.

  

1 2 1 0 0 0 1 1 0 1 0 0 1 1 0 0 1 0

  

  

1 0 − 1 0 2 0 0 1 1 0 − 1 0 0 0 0 1 − 1 0

   Answer: Add s 2 which gives the basis {v 1 , v 2 , s 2 }. C OROLLARY. Suppose V is a vector space. D If W is a maximal set of independent vectors (i.e., it can't be extended to a larger set of independent vectors) then W is a basis. E If W is a minimal set of vectors which span V (i.e., no proper subset of W spans V) then W is a basis.

P ROOF. D By the second theorem, W can be extended to a basis. By hypothesis, no strictly larger set is independent. Hence it must be a basis. E By the first theorem, a subset of W is a basis. By hypothesus, no strictly smaller set spans V. Hence W is a basis. E THEOREM. Suppose U and W are subsets of a vector space V and suppose U has u elements and W has w elements. D If U is independent and W spans V, then u  w. E If U and W are both bases, then u = w. P ROOF OF (E FROM D. If U and W are both bases, then U is independent and W spans V, hence by D u  w. Similarly w  u and hence u = w. E DEFINITION. The dimension n of a vector space V is the number of elements in any basis of V. If V = {0}, the dimension is 0. C OROLLARY. Suppose W is a subset of a vector space V, W has w elements and V has dimension n. D W independent implies w  n ; w > n implies W is dependent. E W spans V implies w! n ; w < n implies W doesn't span V. F W independent and w = n implies W is a basis. G W spans V and w = n implies W is a basis. P ROOF. D and E follow the the theorem above. F and G follow from the corollary above. E C{t 3 , t^2 , t, 1} is a basis for P 3 , the vector space of polynomials of degree 3. Thus P 3 has dimension 4. Explain why each the following cannot be a basis for P 3. {t^3 +1, t^2 +1, t+1} {t 3 +1, t^3 1, t^2 , t, 1} Page 102 Hw 9 Answers 4(2). (c+d)€u = c€u  d€u fails. 6(2). u-u = 0 for some -u fails. 8(7). V = {r>0}, uv = uv (ordinary multiplication), r€u = u r^. (1) Prove uv = vu. uv = uv = vu = vu (2) u(vw) = u(vw) =u(vw) = (uv)w = (uv)w= (uv)w (3) 0 =1. Prove u 0 = u: u 0 = u1 = u. (4) - u = 1/u. Prove u-u = 0: u-u = u(1/u) = 1 = 0 (6) (c+d)€u = u(c+d)^ = (u c)(u d^ ) =(c€u)(d€u) = c€u  d€u (7) c€(d€u) =(d€u) c^ = (u d) c^ =udc^ =(cd)€u. (8) 1€u = u 1 =u 10(7). V = {0}. 00 = 0, r€0 = 0. (1) uv = 00 = 0 = 00 = vu (3) 0 = 0. Prove u 0 = u: u 0 = 00 = 0 = u (4) - u = u. Prove u-u = 0 : u-u = 00 = 0 (5) c€(uv) = c€(00) = c€0 = 0 = 00 = c€uc€v (6) (c+d)€u = (c+d)€0 = 0 = 00 = c€ud€u (7) c€(d€u) = c€0 = 0 = (cd)€u (8)1€u = 0 = u 12(2). V = {xLR: x>0} uv = uv1, c€u = u. The following fail: u(vw) = (uv) w, For some 0, u 0 = u , u-u = 0 for some -u, (c+d)€u=c€ud€u 14(2). V = R, uv = 2uv, c€u = cu. The following fail: uv = vu, u(vw) = (uv)w For some 0, u 0 = u, u-u = 0 for some -u, (c+d)€u=c€ud€u