Dr. Neal, WKU
Linear Combinations in Rn
MATH 307
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n
Let u1, u2 , . . ., um be a collection of m distinct vectors in R . Given another vector v in
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Rn , we wish to test whether or not v is a linear combination of u1, u2 , . . ., um . That is, do
there exist scalars c1 , c2 , . . ., cm such that
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€
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v = c1 u1 + c 2 u2 + . . . + c m um ?
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n
Each vector ui has n coordinates in R , so we may write
€ € 
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u1 = ( u11 , u12 , . . ., u1n )
u2 = ( u21 , u2 2 , . . ., u2 n ) . . . . . um = ( um1 , um 2 , . . ., um n ).
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Likewise we may write v = ( v1, v2 , . . ., vn ).
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In order for v to be a linear combination of u1, u2 , . . ., um , then for some scalars c1 ,
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have c1 u1 + c 2 u2 + . . . + c m um = v . Thus,
c2 , . . ., cm we must
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€
. . ., u2 n )€+ . . . + cm ( um1 , um 2 , . . ., um n )
c1 ( u11 , u12 , . . ., u1n ) + c2 ( u21 , u2€2 , €
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€
= ( v1, v2 , . . ., vn ).
Adding componentwise we obtain
c1 u11 + c2 u21 + . . . +
c1 u12 + c2 u2 2 + . . . +
.
.
c1 u1n + c2 u2 n + . . . +
cm um1 = v1
cm um 2 = v2
cm um n = vn
We thereby obtain a system of n equations and m unknowns.
The m unknowns are the scalars c1 , c2 , . . ., cm . The associated augmented matrix
n
comes from the coordinates of the vectors in R where the vectors are written in column
form:
€

u1

u2
. . .
 u11

 u12
 .

 .
u
 1n
u21
u2 2
.
.
u2 n
.
.
.
.
.
.
.
.
.
.

um

v
. um1 v1 

. um 2 v 2 
.
.
.

.
.
.
. um n v n 
Dr. Neal, WKU
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• If there is no solution to this system, then v is not a linear combination of the vectors
u1, u2 , . . ., um .
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• If there is a solution to the system, then v is a linear combination of u1, u2 , . . ., um
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€ €
m
€ we can find scalars c , c , . . ., c such that v = ∑ ci ui .
and
m
1 2
€
i =1
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3
Example 1. In R , let u1 = (1, –2, 4), u2€= (2, –4, 6), and
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vector v = (2, 4, 6) is a linear combination of u1, u2 , and
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u = (–1, 3, 8). Determine if the
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u3. If so, then write it as such.
Solution. We check to see if there is a solution to the associated system of equations:
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 € €
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€
u1 u2 u3 v
 1 0 0 −88


 1 2 −1 2 

 → rref →  0 1 0 49 


−2 −4 3 4 
0 0 1 8 
 4 6 8 6


 
Thus, there is a solution to the system and (2, 4, 6) is a linear combination of u1, u2 ,
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and u3. We see that (2, 4, 6) = –88 (1, –2, 4) + 49 (2, –4, 6) + 8 (–1, 3, 8), or in simpler
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notation, v = −88
u1 + 49 u2 + 8 u3 .
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€ €
n
Note: When we have n vectors in R , then we can write them as columns to form an
n × n matrix of coefficients A . If det (A) ≠ 0, then every system of equations AX = B will
€
n
have a unique solution. Thus, any other vector in R will be a linear combination of
these first n vectors.
 1 2 −1


In the preceding example, A =  −2 −4 3  and det (A) = 2 ≠ 0. Thus every vector

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 4 6 8
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3
in R is some linear combination of the three vectors u1 = (1, –2, 4), u2 = (2, –4, 6), and
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u3 = (–1, 3, 8).
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4
Example 2. In R , let u1 = (1, –2, 4, 2), u2 = (2, –4, 6, 3), and u3 = (–1, 3, 8, 2). Determine
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if v = (2, 4, 6, 8) is a linear combination of u1, u2 , and u3.
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Solution. We check to see if there is a solution to the associated system of equations:
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€
€
€
1
 1 0 0 0
2 €−1 €2




 −2 −4 3 4
 0 1 0 0

 → rref → 

6
8 6
4
 0 0 1 0
2
 0 0 0 1
3 2 8



Dr. Neal, WKU
no solution to the system, then
 The last row gives an inconsistency. Because
  there is
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v = (2, 4, 6, 8) is not a linear combination of u1, u2 , and u3.
Because we must solve a system of equations in order to determine whether or not a
vector is a linear combination of other vectors, we may not always have a unique
€ €possibilities.
€
solution. There may in fact be infinite
If there are infinitely many possible
linear combinations, then we can still give one particular such solution.
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4 
Example 3. In R , u1 = (1, –2, 4, 2), u2 = (2, –4, 6, 3), and u3 = (5, –10, 16, 8). Determine
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if v = (–2, 4, –2, –1) is a linear combination of u1, u2 , and u3.
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Solution. Consider
the associated
system of equations:
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€
1
1
2
5 −2



 −2 −4 −10 4 
0

 → rref → 
6
16 −2
4
0
2

0
3
8 −1


0
1
0
0
1
2
0
0
4

−3

0
0 
We see that there are infinitely many solutions:
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v = c1 u1 + c 2 u2 + c 3 u3 , where
c1 = 4 − t ; c2 = −3 − 2t ; c3 = t , and t can be any real number.
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€ 4, –2, –1) can be written as a linear combination of u , u , and u
So the vector v = (–2,
1 2
3
in infinitely many ways. One particular solution (when t = 1) is c1 = 3, c 2 = –5, c3 = 1.
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Thus, v = 3 u1 – 5 u2 + u3.
However
the simplest solution is when t = 0. Then we have c1 = 4 and c2 = –3.
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Thus, v = 4 u1 – 3 u2 .
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Note that the vectors u1 and u2 reduce to the 2 × 2 identity after rref. Thus, any
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€ that
€ can
€ be written as linear combination of u , u , and u can actually be written
vector
1 2
3
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as
a
linear
combination
of
just
and
.
u
u
1
2
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€
Dr. Neal, WKU
Exercises
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3
1. In R , let u1 = (1, –2, –1)
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u2 = (–1, 3, 2)
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u3 = (2, –4, –2)
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u4 = (0, 2, 2)
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Let v = (11, 24, 10) and w = (11, –27, –16).
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w are linear€combinations of€u1, u2 , u3, and u4 . If so, give the
(a) €
Determine if v and
general form of all possible linear combinations as well as a specific linear combination.
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€
(b) Which are the
only vectors really necessary to write any other vector as a linear
€ of €u , u ,  , and u ?
€ € €
€
combination
1 2 u3
4
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2. In R 4 , let u1 = (1, –2, –1, 2)
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€
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u5 = (1, –1, 0, 1)
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u2 = (–1, 3, 2, 3)
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u3 = (2, –4, –2, 4)
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u4 = (0, 2, 2, 0) and
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Let v = (1, 2, 3, 4) and w = (4, 3, 2, 1).
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€
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(a) Determine if v and w are linear combinations of u1, u2 , u3, u4 , and u5. If so, give
the general form of all possible linear combinations as well as a specific linear
€
€
combination.
€ are €the only vectors really necessary
€ € to €write
€ any other
€
(b) Which
vector as a linear
   

combination of u1, u2 , u3, u4 , and u5?
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