LINEAR COMBINATION MATH15-1
2nd Quarter AY 2014-2015
LINEAR COMBINATION If w is a vector in a vector space V, then w is said to be a linear combination of the vectors , , , ⋯ , in V if w can be expressed in the form
= + + + ⋯ + Where , , , ⋯ , are scalars. These scalars are called the coefficients of the linear combination. combination.
Theorem: If S = , , … is a nonempty set of vectors in a vector space V, then: (a)The set W of all possible linear combinations of the vectors in S is a subspace of V. (b)The set W in part (a) is the “smallest “ subspace of V that contains all of the vectors in S in the sense that any other subspace that contains those vectors contains W.
EXAMPLE In let = 1, 2, 1 , = 1, 0, 2 , = 1, 1, 0 . The vector = (2, 1, 5) is a linear combination of , , if we can find real numbers , , so that + + = . Substituting , , ,
we have
1, 2, 1 + 1, 0, 2 + 1, 1, 0 = 2, 1, 5 . Combining terms on the left and equating corresponding entries leads to the linear system
+ + = 2 2
+ = 1
+ 2
= 5.
Solving this linear system gives = 1, = 2,
= −1, which means that v is a linear combination of , , . Thus,
= + 2 −
Examples 1. Write = (2, −5, 3) as a linear combination of = 1, −3, 2 , = 2, −4, −1 , and
= 1, −5, 7 . 2. Write the polynomial = 4 − 2 + 5 as a linear combination of the polynomials = 2 + + 1, = − 2 + 2 and
= + 3 + 6 .
Examples 3. Which of the following are linear combinations of u= (1,-3,2) and v= (1,0,-4) (a) (0,-3,6)
(b) (3, -9,-2)
(c) (0,0,0) (d) (1,-6,-16)
4. Which of the following are linear combinations of A=
2 (a) −2
3 0
2 , 1
B=
5 4 (b) 4 −2
0 −2
1 2 , C= −2 4
1 5 (c) −4 10
1 ? 5
9 3 (d) −8 1
9 21
Example 5. Consider the vectors u = (1,2,-1), and v= (6,4,2) in . Show that w = (9, 2,7) is a linear combination of u and v and that w’ =( 4,-1,8) is not a linear combination of u and v?
Span of S The subspace of a vector space V that is formed from all possible linear combinations of the vectors in a nonempty set S is called the span of S, and we say that the vectors in S span that subspace. If S = , , … , then we denote the span of S by span , , … or span (S).
In other words, if S = , , … is a set of vectors in a vector space V, then the set of all vectors in V that are linear combinations of the vectors in S is denoted by span , , … or span (S) The vectors , , ⋯ , in a vector space V are said to span V if every vector in V is a linear combination of , , ⋯ ,
The
procedure to check if the vectors , , ⋯ , pan the vector space V is as follows. Step 1. Choose an arbitrary vector v in V. Step 2. Determine if v is a linear combination of the given vectors. If it is, then the given vectors span V. If it is not, they do not span V.
Examples 1. In let = 2 + + 2, = − 2, = 5 − 5 + 2, = − − 3 − 2. Determine if the vector = + + 2 belongs to span , , , . 2. Determine whether the following polynomials span .
= 1 − + 2 ,
= 3 + ,
= 5 − + 4 ,
= − − +
Examples 3. Let V be the vector space and let
= 1, 2, 1 , = 1, 0, 2 , = (1, 1, 0). Do , , span V?
1 0 0 1 0 0 , , 4. Show that = spans 0 0 1 0 0 1 the subspace of consisting of all symmetric matrices. 5. Let V be the vector space . Let = , () , where = + 2 + 1 and = + 2. Does S span ?
LINEAR INDEPENDENCE If S = , , ⋯ , is a nonempty set of vectors in a vector space V, then the vector equation
+ + ⋯ + = 0 has at least one solution namely, = 0, = 0,…, =0 . We call this the trivial solution. If this is the only solution, then S is said to be a linearly independent set. If there are solutions in addition to the trivial solution, then S is said to be a linearly dependent set.
The procedure to determine if the vectors , , ⋯ , are linearly dependent or linearly independent is as follows. Step 1. Form the equation + + ⋯ + = 0 which leads to a homogeneous system. Step 2. If the homogeneous system obtained in Step 1 has only the trivial solution, then the given vectors are linearly independent; if it has a nontrivial solution, then the vectors are linearly dependent.
Very Important Note
Every set of vectors containing the zero vector is linearly dependent.
Examples 1. Are the vectors = 1, 0, 1, 2 , = (0, 1, 1, 2) and = (1, 1, 1, 3) in linearly dependent or linearly independent? 2. Consider the vectors = 1, 2, −1 , = 1, −2, 1 , = (−3, 2, −1) and = (2, 0, 0) in . Is = , , , linearly dependent or linearly independent?
Examples 3. Consider the vectors = + + 2 ,
= 2 + , = 3 + 2 + 2 . Is = , , () linearly dependent or linearly independent.
BASIS The vectors , , ⋯ , in a vector space V are said to form a basis for V if
a) , , ⋯ , span V b) , , ⋯ , are linearly independent If , , ⋯ , form a basis for vector space V, then they must be distinct and nonzero, hence is written as a set , , ⋯ , .
Examples 1. The vectors = (1, 0) and = (0, 1) form a basis for , the vectors , , and form a basis for and in general, the vectors , , ⋯ , form a basis for . Each of these sets of vectors is called the natural basis or standard basis for , , and , respectively. 2. Show that the set = , , , , where
= 1, 0, 1, 0 , = 0, 1, −1, 2 , = 0, 2, 2, 1 , and = 1, 0 0, 1 is a basis for .
Examples 3. Show that the set = + 1 , − 1 , 2 + 2 is a basis for the vector space . 4. Find a basis for the subspace V of , consisting of all vectors of the form + + , where = − . 5. Let = , , , , be a set of vectors in , where = 1, 2, −2, 1 , =
−3, 0, −4, 3 , = 2, 1, 1, −1 , = −3, 3, −9, 6 , = (9, 3, 7, −6). Find a subset of S that is a basis for = span S.
Let = and let = , , ⋯ , be a set of nonzero vectors in V. The procedure for finding the subset for S that is the basis for = is as follows. Step 1. Form the equation + + ⋯ + = 0. Step 2. Construct the augmented matrix associated with the homogeneous system of equation in step 1 and transform it to RREF. Step 3. The vectors corresponding to the columns containing the leading 1’s form a basis for =
Dimension The dimension of a nonzero vector space V is the number of vectors in a basis for V. We often write dim V for the dimension of V. Since the set 0 is linearly dependent, it is natural to say that 0 has dimension zero.
Examples 1. The dimension of is 2; the dimension of is 3; and in general, the dimension of is n. 2. The dimension of is 3; the dimension of is 4; and in general, the dimension of is + 1. 3. The subspace of W given in Example 5 of the preceding topic is 2.
HOMOGENEOUS SYSTEMS The procedure for finding a basis for the solution space of a homogeneous system = 0, or the null space of A, where A is is as follows. Step 1. Solve the given homogeneous system by Gauss-Jordan reduction. If the solution contains no arbitrary constants, then the solution space is 0 , which has no basis; the dimension of the solution space is zero.
Step 2. If the solution x contains arbitrary constants, write x as a linear combination of vectors , , ⋯ , with , , ⋯ , as coefficients:
= + + ⋯ + . Step 3. The set of vectors , , ⋯ , is a basis for the solution space of = 0; the dimension of the solution space is .
Example 1. Find a basis for and the dimension of the solution space W of the homogeneous system
1 1 0 1 0 0 1−1 2 1
4 2 0 0 6
1 2 1 1 1 2 0 2 0 1
0 0 = 0 0 0
