MTH501 Final Term Highlighted Handouts

Lecture 22

Linearly Independent Sets; Bases

First we revise some definitions and theorems from the Vector Space:

Definition Let V be an arbitrary nonempty set of objects on which two operations are defined, addition and multiplication by scalars.

If the following axioms are satisfied by all objects u, v, w in V and all scalars l and m, then we call V a vector space.

Axioms of Vector Space

For any set of vectors u, v, w in V and scalars l, m, n: 1. u + v is in V 2. u + v = v + u 3. u + (v + w) = (u + v) + w

4. There exist a zero vector 0 such that 0 + u = u + 0 = u

5. There exist a vector – u in V such that -u + u = 0 = u + (-u) 6. (l u) is in V 7. l (u + v)= l u + l v 8. m (n u) = (m n) u = n (m u) 9. (l +m) u= I u+ m u 10. 1u = u where 1 is the multiplicative identity

Definition A subset W of a vector space V is called a subspace of V if W itself is a vector space under the addition and scalar multiplication defined on V.

Theorem If W is a set of one or more vectors from a vector space V, then W is subspace of V if and only if the following conditions hold:

(a) If u and v are vectors in W, then u + v is in W (b) If k is any scalar and u is any vector in W, then k u is in W.

Definition The null space of an m x n matrix A (Nul A) is the set of all solutions of the hom equation Ax = 0 Nul A = {x: x is in Rn and Ax = 0}

Definition The column space of an m x n matrix A (Col A) is the set of all linear combinations of the columns of A. If A = [a1 … an], then Col A = Span { a1,… , an } agr ye 10 Axioms follow kry jo usy hi khy gy ye vector-Space hau.... Sub-Space => agr kisi vector ka subset ly or wo khud vector space bn jye to usy sub-space khy gy

Sub-Space check krny k liye : agr ksii ko check krna ho vector space hai k nei us k liye 10 Axioms check krni prti hai lakin ye ak hi theorm se verify ho jata h is me 2 condition chck krni hotii hai Since we know that a set of vectors ,... p S v v v v spans a given vector space V if every vector in V is expressible as a linear combination of the vectors in S. In general there may be more than one way to express a vector in V as linear combination of vectors in a spanning set. We shall study conditions under which each vector in V is expressible as a linear combination of the spanning vectors in exactly one way. Spanning sets with this property play a fundamental role in the study of vector spaces. In this Lecture, we shall identify and study the subspace H as “efficiently” as possible. The key idea is that of linear independence, defined as in Rn.

Definition An indexed set of vectors {v1,…, vp} in V is said to be linearly independent if the vector equation

p c +c +...+c = 0

p · v · v · v

(1) has only the trivial solution, i.e. c1 = 0, … , cp = 0. The set {v1,…,vp} is said to be linearly dependent if (1) has a nontrivial solution, that is, if there are some weights, c1,…,cp, not all zero, such that (1) holds. In such a case, (1) is called a linear dependence relation among v1, … , vp. Alternatively, to say that the v’s are linearly dependent is to say that the zero vector 0 can be expressed as a nontrivial linear combination of the v’s.

If the trivial solution is the only solution to this equation then the vectors in the set are called linearly independent and the set is called a linearly independent set. If there is another solution then the vectors in the set are called linearly dependent and the set is called a linearly dependent set.

Just as in Rn, a set containing a single vector v is linearly independent if and only if

v

0 . Also, a set of two vectors is linearly dependent if and only if one of the vectors is a multiple of the other. And any set containing the zero-vector is linearly dependent.

Determining whether a set of vectors n a a a a is linearly independent is easy when one of the vectors is 0: if, say, a = , then we have a simple solution to

n n x a x a x a x a given by choosing 1x to be any nonzero value and putting all the other x’s equal to 0. Consequently, if a set of vectors contains the zero vector, it must always be linearly dependent. Equivalently, any set of linearly independent vectors cannot contain the zero vector.

Another situation in which it is easy to determine linear independence is when there are more vectors in the set than entries in the vectors. If n > m, then the n vectors

n a a a a in Rm are columns of an m n × matrix A. The vector equation

n n x a x a x a x a is equivalent to the matrix equation Ax = 0 whose corresponding linear system has more variables than equations. Thus there must be at least one free variable in the solution, meaning that there are nontrivial solutions index-set = wo set jis k sath index value ho jaisyy V1,V2,V3 etc

agr one solution ho or wo trivial ho to wo independent ho ga agr multiple solution hai then its called independent solution to n n x a x a x a x a : If n > m, then the set { n a a a a of vectors in Rm

must be linearly dependent. When n is small we have a clear geometric picture of the relation amongst linearly independent vectors. For instance, the case n = 1 produces the equation x a = , and as long as a ≠ , we only have the trivial solution x = . A single nonzero vector always forms a linearly independent set. When n = 2, the equation takes the form x a x a . If this were a linear dependence relation, then one of the x’s, say 1x , would have to be nonzero. Then we could solve the equation for 1a and obtain a relation indicating that 1a is a scalar multiple of 2a . Conversely, if one of the vectors is a scalar multiple of the other, we can express this in the form x a x a . Thus, a set of two nonzero vectors is linearly dependent if and only if they are scalar multiples of each other.

Example (linearly independent set)

Show that the following vectors are linearly independent:

1 , 1 , v v v

Solution Let there exist scalars c c c in R such that c v c v c v Therefore,

c c c

c c c c c c c

c c c c c c c The above can be written as:

........(1) 0........(4)( (1)) .......(2) 0 ......(3) c c c c dividing by onboth sides of c c c c c