Recall from Section 1 .2 that we defined a basis of a subspace S of JR.11 to be a linearly independent set that spans S. Thus, with our previous tools, we can now easily identify a basis for a subspace. In particular, to show that a set 13 of vectors in JR.11 is a basis for a subspace S, we just need to show that Span 13 = S and 13 is linearly independent. We demonstrate this with a couple examples.
1£1 s =
{[i l
·Hl 'nl}
Prove iliat sis a basis for R3.Solution: To show that every vector v E JR.3 can be written as a linear combination of the vectors in 13, we just need to show that the system
" [ i l
+ 1,Hl
+ 1,[-�l
= vis consistent for all v E JR.3.
Row reducing the coefficient matrix that corresponds to this system to RREF gives
[
2� _;
2 -�1
1 �[� � �1
0 0 1The rank of the matrix is 3, which equals the number of rows (equations). Hence, by Theorem 2.2.2, the system is consistent for all v E JR.3, as required.
Moreover, to determine whether 13 is linearly independent, we would perform the same elementary row operations on the same coefficient matrix. So, we see that the rank of the matrix also equals the number of columns (variables). By Theorem 2.2.2, the system has a unique solution, and hence 13 is also linearly independent. Thus, 13 is a basis for JR.3.
Show that S =
{[
_�l
·[l l}
is a basis for the plane -3x1 + 2x2 + x3 = O.Solution: We first observe that 13 is clearly linearly independent since neither vector is a scalar multiple of the other. Thus, we need to show that every vector in the plane can be written as a linear combination of the vectors in 13. To do this, observe that any vector x in the plane must satisfy the condition of the plane. Hence, every vector in the plane has the form
x =
[
3x1 - 2x2�� l
since X3 = 3x1
-
2x2. Therefore, we now just need to show that the equationEXAMPLE 8
(continued)Theorem 5
Lemma6
is always consistent. Row reducing the corresponding augmented matrix gives
So, the system is consistent and hence :B is a basis for the plane.
A set of vectors {V 1, ... , v n} is a basis for ]Rn if and only if the rank of the coefficient matrix of tiV1 + · · · + tnVn =vis n.
Proof: If {V 1, ... , Vn} is a basis for JRll, then it is linearly independent. Hence, by Lemma 3, the rank of the coefficient matrix is n.
If the rank of the coefficient matrix is n, then the set is linearly independent and
spans ]Rn by Lemma 1 and Lemma 3. •
Theorem 5 gives us a condition to test whether a set of n vectors in JR11 is a basis for 1Rn. Moreover, Lemma 1 and Lemma 3 tell us that a basis of JRll must contain n
vectors. We now want to prove that every basis of a subspace S of ]Rn must contain the same number of vectors.
Suppose that S is a non-trivial subspace of JR11 and Span {vi, ... , ve} = S. If { z11, ... , z1 d is a linearly independent set of vectors in S, then k ::; e.
Proof: Since each z1i, 1 ::; i ::; k, is a vector in S, by definition of spanning, it can be written as a linear combination of the vectors v 1, ... , Ve. We get
Consider the equation
il1 = a11v1 + a11V2 + · · · + ae1Ve z12 = a12V1 + a12V2 + · · · + aeive
_, _, _,
O=t1U1+ .. ·+tkUk
= t1(a11V1 + a11V2 + · · · + ae1Ve) + · · · + tk(alkv1 + a1kV2 + · · · + aekve)
= (a11t1 + · · · + alktkW1 +···+(anti+···+ aektdve
Since {V 1, • • • , v e} is linearly independent, the only solution to this equation is
Theorem 7
Definition
Dimension
EXAMPLE9
Section 2.3 Application to Spanning and Linear Independence 99
This gives a homogeneous system of e equations in the k unknowns t1, ... , tk. If k > e, then this system would have a non-trivial solution, which would imply that {it 1, • • • , uk}
is linearly dependent. But we assumed that {it 1, ... , uk} is linearly independent, so we
must have k � e. •
If {V1, ... , v e} and {it 1, • . . , uk} are both bases of a non-trivial subspace S of IR.n, then k = e.
Proof: We know that {V 1, • • • , v e} is a basis for S, so it is linearly independent. Also, {it 1,. . ., uk} is a basis for S, so Span{u 1, ... , uk} = S. Thus, by Lemma 6, we get e � k.
Similarly, {it j, ... 'uk} is linearly independent as it is a basis for s' and Span{V1' ... 'Ve}
= S, so e � k. Therefore, e = k, as required. •
This theorem justifies the following definition.
If S is a non-trivial subspace of JR.I! with a basis containing k vectors, then we say that the dimension of S is k and write
dims= k
So that we can talk about the basis of any subspace of IR.n, we define the empty set
...,
to be a basis for the trivial subspace {O} of JR.11 and thus say that the dimension of the trivial vector space is 0.
By definition, a plane in JR.11 that passes through the origin is spanned by a set of two linearly independent vectors. Thus, such a plane is 2-dimensional since every basis of the plane will have two vectors. This result fits with our geometric understanding of a plane.
PROBLEMS 2.3
Practice Problems
Al Let B =
{ 1
·! , -� } For each of the following vectors, either express it as a linear combination of the vectors of B or show that it is not a vector in
lowing vectors, either express it as a linear combi
nation of the vectors of B or show that it is not a
neous system that defines the given set.
(a) Span
{[�]
·[!]}
all linear combinations of the vectors that equal the zero vector.A 7 Determine whether the given set is a basis for JR'.3.
vectors, either express it as a linear combination of the vectors of B or show that it is not a vector in
vectors, either express it as a linear combination of the vectors of B or show that it is not a vector in
neous system that defines the given set.
(a) span
{[
-iJ
.lm
all linear combinations of the vectors that are0.
(a)
u, �, n
B6 Determine all values of k such that the given set is linearly independent.
(a)
p , �� , ; }
(b)
h -1.
-; }
Computer Problems
2 1 3 1 0
-1 -2 0 -1 0 1
I 1 -1 0 1 0
Cl Let B = 2 0 5 -4 -2 and v = 0.
4 6 -1 3 0
-2 2 8 6 3 0
3 2 3 0
Conceptual Problems
Dl Let B = {e1, ... ,e11} be the standard basis for JR.11•
Prove that Span B = JR" and that B is linearly inde
pendent.
D2 Let B = {v1, ... 'vd be vectors in JR.11•
B7 Determine whether the given set is a basis for JR.3.
(a)
{[-illlUl}
(b)
mini·m·[�J}
(c)
WlHHm
(d)
rnHm
(a) Determine whether B is linearly independent or dependent.
(b) Determine whether vis in Span B.
(a) Prove that if k < n, then there exists a vector v E JR.11 such that v 'I. Span B.
(b) Prove that if k > n, then B must be linearly dependent.
(c) Prove that if k = n, then SpanB = JR.11 if and only if B is linearly independent.