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Lecture 9
Andrei Antonenko
February 21, 2003
1 Linear combinations
Definition 1.1. Let V be a vector space. A vector v ∈ V is a linear combination of vectors u 1 , u 2 ,... , un if there exist a 1 , a 2 ,... , an ∈ k such that
v = a 1 u 1 + a 2 u 2 + · · · + anun. (1)
Sometimes it is possible to express a vector as a linear combination of other vectors. It can be done by solving a corresponding linear system. We’ll demonstrate it in the following example.
Example 1.2. Consider the space R^2 — the space of all pairs of numbers. Let v = (8, 13), u 1 = (1, 2), and u 2 = (2, 3). Let’s express v as a linear combination of u 1 and u 2. To do this we have to find a and b such that v = au 1 +bu 2 , i.e. (8, 13) = a(1, 2)+b(2, 3) = (a·1+b· 2 , a·2+b·3). So, we get the following system: { 1 a + 2 b = 8 2 a + 3 b = 13
We can simply solve this system: subtracting the first equation multiplied by 2 from the second one we get −b = − 3 , so b = 3, and so a = 8− 2 b = 2. So we see that (8, 13) = 2·(1, 2)+3·(2, 3).
Example 1.3. Consider the space P (t) — space of all polynomials. Let v = 5t^2 + 2t + 1, u 1 = t^2 + t, u 2 = t + 1, u 3 = t^2 + 1. Let’s express v as a linear combination of u 1 , u 2 and u 3. We should find a, b and c such that v = au 1 + bu 2 + cu 3 , i.e. 5 t^2 + 2t + 1 = a(t^2 + t) + b(t + 1) + c(t^2 + 1) = t^2 (a + c) + t(a + b) + (b + c). So, we get the following system:
a + c = 5 a + b = 2 b + c = 1
Let’s solve this system. Subtracting the first equation from the second one we get
a + c = 5 b − c = − 3 b + c = 1
and than subtracting the second from the third one we get
a + c = 5 b − c = − 3 2 c = 4
So, c = 2, b = −3 + 2 = − 1 , and a = 5 − 2 = 3. So, 5 t^2 + 2t + 1 = 3(t^2 + t) − (t + 1) + 2(t^2 + 1).
2 Linear dependence and independence
Now we’ll study one of the most important concepts of linear algebra and the theory of vector spaces. This is a concept of linear dependence and independence.
Definition 2.1. Let u 1 , u 2 ,... , un be a system of vectors. A linear combination of them is called nontrivial if there exists a nonzero coefficient. If all coefficients are equal to 0, the linear combination is called trivial.
Example 2.2. u 1 + 0u 2 + 0u 3 − 3 u 4 is nontrivial linear combination, and 0 u 1 + 0u 2 + 0u 3 + 0u 4 is a trivial linear combination.
Definition 2.3. A system of vectors u 1 , u 2 ,... , un is called linearly dependent if there exists a nontrivial linear combination of these vectors which is equal to zero. Otherwise the system is called linearly independent.
Example 2.4. Consider a vector space R^3. Let u 1 = (3, − 5 , 0), u 2 = (5, 0 , 1), and u 3 = (8, − 5 , 1). Then linear combination with coefficients 1,1, and -1 is nontrivial and equals to zero: 1 · (3, 5 , 0) + 1 · (5, 0 , 1) + (−1) · (8, − 5 , 1) = (0, 0 , 0).
Example 2.5. Consider a vector space R^2. Let u 1 = (1, 1), and u 2 = (0, 0). The linear combination with coefficients 0 and 1 is nontrivial, and equals to zero:
0 · (1, 1) + 1 · (0, 0) = (0, 0)
Moreover, if one of the vectors in the system equals to 0 , then this system is linearly dependent, since we can make a coefficient before it equal to some nonzero number, and all other coefficients we can make equal to zero.
so, u 1 is expressed as a linear combination of other vectors. On the contrary, let one of these vectors, say, u 1 , can be expressed as a linear combination of other vectors: u 1 = b 2 u 2 + b 3 u 3 + · · · + bnun.
Then u 1 − b 2 u 2 − b 3 u 3 − · · · − bnun = 0
is a nontrivial linear combination which is equal to zero, and so vectors are linearly dependent.
Example 2.9. The vectors u 1 = (0, 3 , 5), u 2 = (− 1 , 4 , 7) and u 3 = (1, 2 , 3) are linearly depen- dent since the linear combination with coefficients -2, 1 and 1 is equal to 0:
−2(0, 3 , 5) + (− 1 , 4 , 7) + (1, 2 , 3) = (0, 0 , 0).
So, vector u 1 can be expressed as a linear combination of u 2 and u 3 :
u 1 = (0, 3 , 5) =^12 (− 1 , 4 , 7) +^12 (1, 2 , 3)
Example 2.10. Let
u 1 =
, u 2 =
, u 3 =
Here u 1 , u 2 and u 3 are linearly independent and none of these vectors can be expressed as a linear combination of other 2 vectors. For example, for u 1 there are no real a and b such that
u 1 =
= au 2 +^ bu 3 =^ a
+ b
