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Lecture 10
Andrei Antonenko
February 24, 2003
1 Meaning of linear dependence and independence
On the last lecture we stated the result that if the system of vectors is linearly dependent, then at least one vector of the. system can be expressed as a linear combination of others. We gave an example how to do it. Now we’ll give an example when it is not possible to express any vector as a linear combination of others.
Example 1.1. 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
2 Spanning sets
Definition 2.1. Let V be a vector space. Vectors v 1 , v 2 ,... , vn are called a spanning set of V if every element of V is a linear combination of v 1 , v 2 ,... , vn. In this case the space V is called a span of these vectors and it is denoted by V = 〈v 1 , v 2 ,... , vn〉
Example 2.2. Consider the vector space R^3. Then vectors v 1 = (1, 0 , 0), v 2 = (0, 1 , 0), and v 3 = (0, 0 , 1) form a spanning set of R^3 , since if u ∈ V equals to (a, b, c), then u = av 1 +bv 2 +cv 3.
Example 2.3. Consider the vector space R^3. Then vectors v 1 = (1, 1 , 1), v 2 = (1, 1 , 0), and v 3 = (1, 0 , 0) form a spanning set of R^3 , since if u ∈ V equals to (a, b, c), then u = cv 1 + (b − c)v 2 + (a − b)v 3. For example, (4, 6 , 1) = 1(1, 1 , 1) + 5(1, 1 , 0) − 2(1, 0 , 0).
Example 2.4. Consider the vector space P (t). Then vectors 1 , t, t^2 , t^3 ,... are a spanning set of P (t) since it is clear that every polynomial can be expressed as a linear combination of these vectors.
Example 2.5. Consider a vector space M 2 , 2 of 2 × 2 -matrices. Then the following matrices
form a spanning set for M 2 , 2 : v 1 =
, v 2 =
, v 3 =
, v 4 =
, since
any
a b c d
= av 1 + bv 2 + cv 3 + dv 4.
3 Homogeneous systems
As we saw in the previous lecture, in order to figure out whether the given vectors are linearly dependent or independent, we need to solve linear system with zeros in the right hand side. This leads us to the following topic.
Definition 3.1. A system of linear equations is called homogeneous if right-hand sides of all its equations are equal to 0.
Example 3.2. The following system is homogeneous: { x 1 + 2 x 2 + x 4 = 0 3 x 1 + x 2 − x 3 + 5 x 4 = 0
The most important fact about homogeneous systems is that it has at least one solution, i.e. zero-solution — the solution where all the variables are equal to 0. As we saw before, for general linear systems there are 3 possible cases — no solutions, 1 solution, and infinitely many solutions. Here for homogeneous systems the first case may not happen. So, we have the following lemma:
Lemma 3.3. The homogeneous system of linear equations may have unique solution or in- finitely many solutions.
Moreover, we can give another statement which tells us when the system has infinitely many solutions.
Theorem 3.4. A homogeneous system with the number of equations less then the number of variables has infinitely many solutions.
Proof. Let’s transpose the system to a row echelon form. We will not get an equation of the form 0x 1 + · + 0xn = b, where b > 0. Moreover, there will be free variables (since all the variables can not be leading — each equation has only one leading variable, so the number of
Example 4.2. Vectors
u 1 =
, u 2 =
, u 3 =
form a spanning set for R^2. To check this one should take arbitrary vector and represent it as a linear combination of u 1 and u 2 : ( a b
= au 1 + bu 2 + 0u 3 = a
Moreover, here we can give many different representations, e.g. ( a b
= au 1 + bu 2 + 0u 3 = (a − 1)
For example, ( 3 5
So, we see that there may be many different spanning sets for a vector space, and they may have different number of vectors. The spanning set in the example 4.1 is better than the spanning set in the example 4.2 since it contains less vectors.
Definition 4.3. The system of vectors from vector space V is called basis if it is
- linearly independent and
- forms a spanning set for V.
Example 4.4. Consider the vector space R^2. The system of vectors
u 1 =
, u 2 =
is a basis. Let’s check it.
- This system forms a spanning set since any vector can be represented as a linear combi- nation of u 1 , u 2 and u 3 : ( a b
= au 1 + bu 2 = a
- This system is independent. Let’s check it. We’ll form a linear combination which is equal to 0, and see that it is trivial:
x
This is equivalent to the following system of linear equations: { x = 0 y = 0 So, we get that x = 0, y = 0. So, this linear combination is trivial, and thus the system of vectors is linearly independent.
Example 4.5. Consider the vector space R^2. The system of vectors
u 1 =
, u 2 =
is a basis. Let’s check it.
- This system forms a spanning set since any vector can be represented as a linear combi- nation of u 1 and u 2 : ( a b
= au 1 + (b − a)u 2 = a
- This system is independent. Let’s check it. We’ll form a linear combination which is equal to 0, and see that it is trivial:
x
This is equivalent to the following system of linear equations: { x = 0 x + y = 0 This system has only one solution x = 0, y = 0. So, this linear combination is trivial, and thus the system of vectors is linearly independent.
Example 4.6. Consider a vector space R^2. The system of vectors
u 1 =
, u 2 =
does not form a basis since it is not a spanning set for R^2. For example we can never represent
vector
as a linear combination of u 1 and u 2.
