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Elements of Vector Algebra

−→ λ 2 −→ λ 3 −→
Indeed, if we assume in formula (4.5), for example, λ 1 ̸= 0, then a 1 = − a 2 − a 3 −
λ 1 λ 1
λ n −→ −→
. . . − a n , i.e. vector a 1 can be expressed as a linear combination of the rest of vectors, but
λ 1
it can’t be done in case when all digits λ 1 = λ 2 = ... = λ n = 0. Thus, if some vectors are linear
dependent, then at least one of them can be expressed as a linear combination of the rest ones
and vice versa.

Theorem 4.5.
−→
−→
Any three vectors a , b , c located in a plane are linear dependent. ⋆
−→

C
M

→ c
−
−
−→
λ 2 c → b + λ 2
− = λ 1
→
a
−→
c
− →
λ 1 b
O . . . . . . . . . . B
− →
b
Figure 4.12 – Linear dependence of vectors in a plane

PROOF. 1. There is a pair of colinear vectors, for example, a and b among these vectors.
− →
− →
−→ −→
−→
−→
−→
−→
Then a = λ · b , because they are colinear, and a = λ · b + 0 · c , meaning vector a is
−→ − → −→ −→ −→
a linear combination of vectors b and c . Therefore, vectors a , b , c are linear dependent.
2. There isn’t any pair of colinear vectors among given ones. Let’s assume that all these vectors
havethesame beginning-pointO ( fig. 4.12). Let’sdrawlinesthroughpointO thatareparallel
−→ −→ −→ −→
to vectors b , c towards crossing in points B and C with lines, that contain vectors b , c .
−−→ −−→ −→ −−→ −→ −→ −→
Obviously, OM = OB +OC. As OB and OC are colinear to vectors b and c correspond-
−→ −→ −→ −→
−→
−→
−→
−→
ingly, then OB = λ 1 · b , OC = λ 2 · c . Thus, a = λ 1 · b + λ 2 · c , i.e. vector a is
−→ −→
−→
−→
−→
a linear combination of vectors b and c . Therefore, vectors a , b , c are linear dependent.
2
This theorem takes place for bigger amount of vectors on a plane.
Theorem 4.6.
−→
− →
For two vectors a and b to be linear independent it is necessary and sufficient to be non-
colinear. ⋆
As follows from Theorems 4.5 and 4.6, the maximum number of linear independent vectors
in a plane equals to two.

Theorem 4.7.
−→ −→
−→
−→
Any four vectors a , b , c and d in space are linear dependent. ⋆

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