Question

5. Prove the Following:

a. Let {v1, . . . , vn} be a finite collection of vectors in a vector space V and suppose that it is not a linearly independent set.

i. Show that one can find a vector w ∈ {v1, . . . , vn} such that w ∈ Span(S) for S := {v1, . . . , vn} \ {w}. Conclude that Span(S) = Span(v1, . . . , vn).

ii. Suppose T ⊂ {v1, . . . , vn} is known to be a linearly independent subset. Argue that the vector w from the previous part can be chosen from the set {v1, . . . , vn} \ T.

b. Let V be a vector space and v ∈ V a vector in it. Argue that the set {v} is a linearly independent set if and only if v 6= ~0. Then use this fact together with part i of part a to prove that if {v1, . . . , vn} is any finite subset of V containing at least one non-zero vector, you can obtain a basis of Span(v1, . . . , vn) by simply discarding some of the vectors vi from the set {v1, . . . , vn}.

c. Suppose {v1, . . . , vn} is a linearly independent set in V and that {w1, . . . , wm} is a spanning set in V.

i. Prove that n ≤ m. Hint: use part ii of part a to argue that, for any r ≤ min(m, n), there is a subset T ⊂ {w1, . . . , wm} of size r such that {v1, . . . , vr , w1, . . . , wm} \ T is a spanning set. Then consider the two possibilities when r = min(m, n).

ii. Conclude that if a vector space has a finite spanning set, then any two bases are finite of equal length. (Necessarily, this means that our notion of dimension from class is well-defined and any vector space with a finite spanning set hence has finite dimension).

Answer 1

Part-A:

Since the set is linearly dependent hence there exists scalars not all zero such that

Take

then

Surely

Since so .

Part-B:

Assume that .

The set is linearly independent because if for any scalar we have

since

and hence it is linearly independent.

In order to obtain the basis we just need to remove those such that is not a linear combination of the other vectors in the given list.

Part-C:

Since from any spanning set we can extract those vectors which are linearly independent so .

Suppose are two basis of a vector space.

Since generates the vector space by Part(i) number of elements in is more than .

Thus cardinality of is greater than .------------(1)

Again

Since too generates the vector space by Part(i) number of elements in is less than .

Thus cardinality of is lesser than .------------(2)

Using (1) and (2) have the same cardinality.

Hence the notion of dimension is well-defined

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