- #1

gutnedawg

- 35

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s ≤ m, and let B be an s × n matrix obtained by choosing s distinct rows of A. Prove

that

rank(B) ≥ r + s − m.

Solution:

Assume that s is the largest amount of distinct rows of A.

r = n-dimNul A

dimNul A= m-s

r=n-m+s

rank(B) ≥ n-m +s + s − m

rank(B) ≥ n+ (-2m +2s)

rank(B) = n- dimNul B

dimNul B = 0 since all rows s are linearly independet

rank(B) = n

n ≥ n -2m +2s

0 ≥ -2m + 2s

2m ≥ 2s

m ≥ s (dividing by 2)

Since this is given doesn't this conclude the proof? Or should I plug in the 0 so

rank(B) ≥ n + x where 0≥x

n ≥ n + x

2. Let V be a vector space, let p ≤ m, and let b1, . . . , bm be vectors in V such that

A = {b1, . . . , bp} is a linearly independent set, while C = {b1, . . . , bm} is a spanning set

for V . Prove that there exists a basis B for V such that A ⊆ B ⊆ C.

Solution:

I'm going on the fact that it does not mention C is linearly independent, thus by the spanning set theorem there exists a linearly independent set of vectors {bi,...,bk} which spans V. Thus, this set {bi,...,bk} is a basis for V.

This means that the basis must at least be equal to A since B cannot be a basis for V if there is another linearly independent vecotr bp. Meaning:

[tex] A \subseteq B [/tex]

Also since B is a spanning set of V and is comprised of at least {b1,...,bp} it must be a subset of C since C also spans V and includes A.

Thus

[tex] A \subseteq B \subseteq C [/tex]

3. Let {v1 , v2, . . . , vm} be an indexed linearly dependent set of vectors in a vector space V such that v1 is not 0v . Prove that there exists exactly one index 2≤ i≤m with the property that the vector vi can be expressed as a linear combination of the preceding vectors v1, . . . , v(i-1) in a unique way.