MCV4U1 – UNIT SIX LESSON SEVEN
Lesson Seven: Linear Combinations and Spanning Sets
Recall that the vectors iˆ, ˆj , and kˆ are called basis vectors. In two dimensions, iˆ and ˆj are called a basis for the
plane, and in three dimensions, iˆ, ˆj , and kˆ are called a basis for space. Or we just use the general term “spanning
sets.”
However, these unit vectors are not the only ones that can make a spanning set. Any two non-collinear/parallel vectors
can be a spanning set in two dimensions.
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Ex. Write u  11,2 as a linear combination of v  1,2 and w  2,1 , thereby showing that 1,2, 2,1 is a
spanning set.
Solution:
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Let u  kv  lw
11,2  k 1,2  l 2,1
If LS=RS, then the individual components must equal too.
11  k  2l
 2  2k  l
 11,2   31,2  42,1
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or u  3v  4w
Ex. Is
Solve to get k  3 and l  4 .
4,2,  6,3 a spanning set?
Solution:
Since
3
4,2   6,3, the vectors are parallel and therefore cannot be a spanning set. (i.e., they don’t define a two
2
dimensional plane, or surface, they only define a line.)
Defines a
line, not a
surface.
Defines a
plane
Now it gets a bit abstract. In three dimensions, any three non-coplanar vectors can be a spanning set.
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Ex. Write u   24,30,47 as a linear combination of v  1,3,2 and w  2,4,5, and x   5,3,7 , thereby
showing that 1,3,2, 2  4,5,  5,3,7 is a spanning set.
Solution:
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Let u  kv  lw  mx
 24,30,47  k 1,3,2  l 2,4,5  m 5,3,7
MCV4U1 – UNIT SIX LESSON SEVEN
If LS=RS, then the individual components must equal too.
Set up a system of three equation in three unknowns, and solve.
 24  k  2l  5m
30  3k  4l  3m
 47  2k  4l  7m
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u  2v  3w  4x and
Spanning
set in three
dimensions.
k  2, l  3, m  4
1,3,2, 2  4,5,  5,3,7 is a spanning set.
Not a
spanning set
in three
dimensions.