Math 311 Lecture 12
DEFINITION. Vectors v1, v2, ..., vn are a basis for a vector
space V iff they span V and they are independent.
CThe standard basis for R3 is . Clearly
{
1
0
0
,
0
1
0
,
0
0
1
}
the three vectors span R3 and are independent.
CThe standard basis for P2 is .
{t2,t,1
}
The standard basis for other spaces is defined similarly.
THEOREM. If v1, v2, ..., vn are a basis for V, then every
vector of V can be written in one and only one way as
a linear combination of v1, v2, ..., vn.
PROOF. Suppose v can be written as two linear
combinations a1v1+a2v2+ ... +anvn and b1v1+b2v2+ ...
+bnvn. Then (a1b1)v1+(a2b2)v2+...+(anbn)vn =
(a1v1+a2v2+ ... +anvn)(b1v1+b2v2+ ... +bnvn) = vv = 0.
By linear independence, (a1b1) = (a2b2) = ... = (anbn)
= 0. Hence the linear combinations must be the
same.E
RECALL. If a set of vectors is dependent, then one can be
written as a linear combination of the others.
THEOREM. If v1, v2, ..., vn span W, then some subset of v1,
v2, ..., vn is a basis for W.
PROOF. If v1, v2, ..., vn are not independent, then one can be
written in terms of the others. Delete this vector. If the
result is still dependent, then again one can be written
in terms of the others, and again delete it. Repeat the
process until the set becomes independent. E
Here is the systematic way to do this.
LEMMA. Reduce the matrix A = (v1_v2_..._vn) to rref. Then
the vectors vi whose columns in the rref matrix have
leading 1's form a basis for the span W of v1, v2, ..., vn.
PROOF. Suppose n = 4. Consider the homogeneous system:
. Suppose that in the rref
x
1v1+x2v2+x3v3+x4v4=0
matrix, just the columns for v1 and v4 have leading 1's.
Then in the general solution, x2 and x3 are arbitrary
variables. We claim that v2 and v3 are dependent.
To write v2 in terms of v1 and v4, set x2 = 1 and x3 = 0. Thus
. Thus .
x
1v1+1v2+0v3+x4v4=0v2=−x1v1−x4v4
To write v2 in terms of v1 and v4, set x2 = 1 and x3 = 0.
Delete the dependent vectors v2 and v3 to get the basis
{v1, v4}. E
With this procedure the deleted vectors will, in fact,
depend only on earlier undeleted vectors. Hence v2 and
v3 will be written in terms of just the earlier vector v1.
CFind a basis for the subspace W spanned by the vectors
v1 = [1,-1,0], v2 = [-1,1,0], v3 = [1,0,1], v4 = [0,1,1].
First convert to the equivalent column vector problem.
Second, eliminate the dependent vectors.
To find the dependencies, solve the homogenous
system of equations: .
x
v1+yv2+zv3+wv4=0
Reducing the augmented matrix to rref gives:
and hence we get
1−1100
−1 1 010
0 0 110
→
1−10−10
00110
00000
.
x
=y+w,z=−w,y,warbitrar
y
By the lemma, the original vectors whose rref columns
have leading 1's are a basis. Converting back to rows
gives:
Answer: v1 = [1,-1,0] and v3 = [1,0,1] are a basis.
To write v2 in terms of v1 and v3, set the variable y for v2
to 1, set w = 0.
Then x = y+w =1+0= 1, z = -w = 0. Hence
Hence = 0 becomes
x
v1+yv2+zv3+wv4
. Solve for v2: . 1v1+1v2+0v3+0v4=0v2=−v1
Likewise v4 can also be written as a linear combination of
v1 and v3. Set w = 1, y = 0. x = y+w =0+1= 1, z = -w =-1.
Hence . So .1v1+0v2+(−
1)v3+1v4=0v4=v3−v1
CFind a basis for the subspace of all skew symmetric
matrices of R 3x3.
An arbitrary such matrix looks like .
0ab
−a0c
−b−c0
Separate the parts with , then factor out a,b,c:
a,b,c
0ab
−a0c
−b−c0
=
0a0
−a00
000
+
00
b
−b00
000
+
000
00c
0−c0
. Hence
a
010
−100
000
+b
001
000
−100
+c
000
001
0−10
is a basis.
{
010
100
000
,
001
000
−100
,
000
001
0−10
}
CIn P3, find a basis for the subspace of polynomials of
the form (a+b)t2+(a+b)t+b+c.
Separating the parts gives that any such polynomial can
be written as the sum
(a+b)t2+(a+b)t+c = (at2+at) + (bt2+bt+a) + (c)
= a(t2+t) + b(t2+t+1) + c(1).
Hence any such polynomial can be written as a sum of the
three polynomials t2+t, t2+t+1, 1.
Are these independent? No, t2+t+1is the sum of t2+t and
1.
Deleting this dependent vector gives {t2+t, 1} which is
independent and hence a basis.
CFind a basis for the subspace V = {[a, b, c] : a = b+c}.
Replacing a by b+c gives [a, b, c] =
[b+c, b, c] = [b, b, 0] + [c, 0, c] = b[1, 1, 0] + c[1, 0, 1].
Hence every vector in V is a linear combination of
[1, 1, 0] and [1, 0, 1].
