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The Replacement Theorem Theorem (Theorem 1.10) Let V be a vector space and suppose G and L are finite subsets of V such that V = Span ( G ) , |G| = n , L is linearly independent, and |L| = m . Then m n and there is a set H G , such that

SLIDE 1

The Replacement Theorem

Theorem (Theorem 1.10) Let V be a vector space and suppose G and L are finite subsets of V such that V = Span(G), |G| = n, L is linearly independent, and |L| = m. Then m ≤ n and there is a set H ⊂ G, such that |H| = n − m and Span(H ∪ L) = V .

SLIDE 2

The Replacement Theorem

Theorem (Theorem 1.10) Let V be a vector space and suppose G and L are finite subsets of V such that V = Span(G), |G| = n, L is linearly independent, and |L| = m. Then m ≤ n and there is a set H ⊂ G, such that |H| = n − m and Span(H ∪ L) = V .

SLIDE 3

The Replacement Theorem

Theorem (Theorem 1.10) Let V be a vector space and suppose G and L are finite subsets of V such that V = Span(G), |G| = n, L is linearly independent, and |L| = m. Then m ≤ n and there is a set H ⊂ G, such that |H| = n − m and Span(H ∪ L) = V .

SLIDE 4

The Replacement Theorem

Theorem (Theorem 1.10) Let V be a vector space and suppose G and L are finite subsets of V such that V = Span(G), |G| = n, L is linearly independent, and |L| = m. Then m ≤ n and there is a set H ⊂ G, such that |H| = n − m and Span(H ∪ L) = V .

SLIDE 5

The Replacement Theorem

Theorem (Theorem 1.10) Let V be a vector space and suppose G and L are finite subsets of V such that V = Span(G), |G| = n, L is linearly independent, and |L| = m. Then m ≤ n and there is a set H ⊂ G, such that |H| = n − m and Span(H ∪ L) = V .

SLIDE 6

Proof

We proceed by induction on m. If m = 0 then L = φ and we let H = G so that m = 0 ≤ n and Span(H ∪ L) = Span(G) = V . Now suppose the statement is true for m = µ, where µ is a fixed nonnegative integer. We assume that L = { v1, v2, . . . , vµ, vµ+1} is linearly independent. Then the set { v1, v2, . . . , vµ} is linearly

independent. So by the induction hypothesis there is a subset

{ u1, u2, . . . , un−µ} ⊂ G such that Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) = V .

SLIDE 7

Proof

We proceed by induction on m. If m = 0 then L = φ and we let H = G so that m = 0 ≤ n and Span(H ∪ L) = Span(G) = V . Now suppose the statement is true for m = µ, where µ is a fixed nonnegative integer. We assume that L = { v1, v2, . . . , vµ, vµ+1} is linearly independent. Then the set { v1, v2, . . . , vµ} is linearly

independent. So by the induction hypothesis there is a subset

{ u1, u2, . . . , un−µ} ⊂ G such that Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) = V .

SLIDE 8

Proof

We proceed by induction on m. If m = 0 then L = φ and we let H = G so that m = 0 ≤ n and Span(H ∪ L) = Span(G) = V . Now suppose the statement is true for m = µ, where µ is a fixed nonnegative integer. We assume that L = { v1, v2, . . . , vµ, vµ+1} is linearly independent. Then the set { v1, v2, . . . , vµ} is linearly

independent. So by the induction hypothesis there is a subset

{ u1, u2, . . . , un−µ} ⊂ G such that Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) = V .

SLIDE 9

Proof

We proceed by induction on m. If m = 0 then L = φ and we let H = G so that m = 0 ≤ n and Span(H ∪ L) = Span(G) = V . Now suppose the statement is true for m = µ, where µ is a fixed nonnegative integer. We assume that L = { v1, v2, . . . , vµ, vµ+1} is linearly independent. Then the set { v1, v2, . . . , vµ} is linearly

independent. So by the induction hypothesis there is a subset

{ u1, u2, . . . , un−µ} ⊂ G such that Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) = V .

SLIDE 10

Proof

We proceed by induction on m. If m = 0 then L = φ and we let H = G so that m = 0 ≤ n and Span(H ∪ L) = Span(G) = V . Now suppose the statement is true for m = µ, where µ is a fixed nonnegative integer. We assume that L = { v1, v2, . . . , vµ, vµ+1} is linearly independent. Then the set { v1, v2, . . . , vµ} is linearly

independent. So by the induction hypothesis there is a subset

{ u1, u2, . . . , un−µ} ⊂ G such that Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) = V .

SLIDE 11

Proof

We proceed by induction on m. If m = 0 then L = φ and we let H = G so that m = 0 ≤ n and Span(H ∪ L) = Span(G) = V . Now suppose the statement is true for m = µ, where µ is a fixed nonnegative integer. We assume that L = { v1, v2, . . . , vµ, vµ+1} is linearly independent. Then the set { v1, v2, . . . , vµ} is linearly

independent. So by the induction hypothesis there is a subset

{ u1, u2, . . . , un−µ} ⊂ G such that Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) = V .

SLIDE 12

Proof

We proceed by induction on m. If m = 0 then L = φ and we let H = G so that m = 0 ≤ n and Span(H ∪ L) = Span(G) = V . Now suppose the statement is true for m = µ, where µ is a fixed nonnegative integer. We assume that L = { v1, v2, . . . , vµ, vµ+1} is linearly independent. Then the set { v1, v2, . . . , vµ} is linearly

independent. So by the induction hypothesis there is a subset

{ u1, u2, . . . , un−µ} ⊂ G such that Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) = V .

SLIDE 13

Proof

We proceed by induction on m. If m = 0 then L = φ and we let H = G so that m = 0 ≤ n and Span(H ∪ L) = Span(G) = V . Now suppose the statement is true for m = µ, where µ is a fixed nonnegative integer. We assume that L = { v1, v2, . . . , vµ, vµ+1} is linearly independent. Then the set { v1, v2, . . . , vµ} is linearly

independent. So by the induction hypothesis there is a subset

{ u1, u2, . . . , un−µ} ⊂ G such that Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) = V .

SLIDE 14

Proof

We proceed by induction on m. If m = 0 then L = φ and we let H = G so that m = 0 ≤ n and Span(H ∪ L) = Span(G) = V . Now suppose the statement is true for m = µ, where µ is a fixed nonnegative integer. We assume that L = { v1, v2, . . . , vµ, vµ+1} is linearly independent. Then the set { v1, v2, . . . , vµ} is linearly

independent. So by the induction hypothesis there is a subset

{ u1, u2, . . . , un−µ} ⊂ G such that Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) = V .

SLIDE 15

Proof

We proceed by induction on m. If m = 0 then L = φ and we let H = G so that m = 0 ≤ n and Span(H ∪ L) = Span(G) = V . Now suppose the statement is true for m = µ, where µ is a fixed nonnegative integer. We assume that L = { v1, v2, . . . , vµ, vµ+1} is linearly independent. Then the set { v1, v2, . . . , vµ} is linearly

independent. So by the induction hypothesis there is a subset

{ u1, u2, . . . , un−µ} ⊂ G such that Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) = V .

SLIDE 16

Proof

We proceed by induction on m. If m = 0 then L = φ and we let H = G so that m = 0 ≤ n and Span(H ∪ L) = Span(G) = V . Now suppose the statement is true for m = µ, where µ is a fixed nonnegative integer. We assume that L = { v1, v2, . . . , vµ, vµ+1} is linearly independent. Then the set { v1, v2, . . . , vµ} is linearly

independent. So by the induction hypothesis there is a subset

{ u1, u2, . . . , un−µ} ⊂ G such that Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) = V .

SLIDE 17

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 18

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 19

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 20

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 21

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 22

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 23

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 24

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 25

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 26

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 27

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 28

Proof (continued)

Hence

vµ+1 = a1

u1 + a2 u2 + · · · + an−µ un−µ + b1 v1 + b1 v2 + · · · + bµ vµ, for some scalars a1, a2, . . . , an−µ, b1, b2, . . . , bµ. We note that n − µ > 0 since otherwise vµ+1 would be a linear combination of

v2, . . . , vµ would contradict L being linearly independent. Therefore n − µ ≥ 1 and n ≥ µ + 1. Similarly at least of the scalars a1, a2, . . . an−µ must be nonzero since L is linearly

independent. Suppose without loss of generality that a1 = 0. Then
u1 = (−a2/a1)

u2 + · · · + (−an−µ/a1) un−µ + (−b1/a1) v1 + · · · + (−bµ)/a1) vµ + (1/a1) vµ+1).

SLIDE 29

Proof (continued)

We let H = { u2, . . . , un−µ}. Then

u1 ∈ Span(H ∪ L),

and { u1, . . . , un−µ, v1, . . . , vµ} ⊂ Span(H ∪ L), V = Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) ⊂ Span(H ∪ L) ⊂ V , and V = Span(H ∪ L), H ⊂ G, |H| = (n − µ) − 1 = n − (µ + 1), and the theorem is true for m = µ + 1. Hence the theorem is true for all m by induction.

SLIDE 30

Proof (continued)

We let H = { u2, . . . , un−µ}. Then

u1 ∈ Span(H ∪ L),

and { u1, . . . , un−µ, v1, . . . , vµ} ⊂ Span(H ∪ L), V = Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) ⊂ Span(H ∪ L) ⊂ V , and V = Span(H ∪ L), H ⊂ G, |H| = (n − µ) − 1 = n − (µ + 1), and the theorem is true for m = µ + 1. Hence the theorem is true for all m by induction.

SLIDE 31

Proof (continued)

We let H = { u2, . . . , un−µ}. Then

u1 ∈ Span(H ∪ L),

and { u1, . . . , un−µ, v1, . . . , vµ} ⊂ Span(H ∪ L), V = Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) ⊂ Span(H ∪ L) ⊂ V , and V = Span(H ∪ L), H ⊂ G, |H| = (n − µ) − 1 = n − (µ + 1), and the theorem is true for m = µ + 1. Hence the theorem is true for all m by induction.

SLIDE 32

Proof (continued)

We let H = { u2, . . . , un−µ}. Then

u1 ∈ Span(H ∪ L),

and { u1, . . . , un−µ, v1, . . . , vµ} ⊂ Span(H ∪ L), V = Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) ⊂ Span(H ∪ L) ⊂ V , and V = Span(H ∪ L), H ⊂ G, |H| = (n − µ) − 1 = n − (µ + 1), and the theorem is true for m = µ + 1. Hence the theorem is true for all m by induction.

SLIDE 33

Proof (continued)

We let H = { u2, . . . , un−µ}. Then

u1 ∈ Span(H ∪ L),

and { u1, . . . , un−µ, v1, . . . , vµ} ⊂ Span(H ∪ L), V = Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) ⊂ Span(H ∪ L) ⊂ V , and V = Span(H ∪ L), H ⊂ G, |H| = (n − µ) − 1 = n − (µ + 1), and the theorem is true for m = µ + 1. Hence the theorem is true for all m by induction.

SLIDE 34

Proof (continued)

We let H = { u2, . . . , un−µ}. Then

u1 ∈ Span(H ∪ L),

and { u1, . . . , un−µ, v1, . . . , vµ} ⊂ Span(H ∪ L), V = Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) ⊂ Span(H ∪ L) ⊂ V , and V = Span(H ∪ L), H ⊂ G, |H| = (n − µ) − 1 = n − (µ + 1), and the theorem is true for m = µ + 1. Hence the theorem is true for all m by induction.

SLIDE 35

Proof (continued)

We let H = { u2, . . . , un−µ}. Then

u1 ∈ Span(H ∪ L),

and { u1, . . . , un−µ, v1, . . . , vµ} ⊂ Span(H ∪ L), V = Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) ⊂ Span(H ∪ L) ⊂ V , and V = Span(H ∪ L), H ⊂ G, |H| = (n − µ) − 1 = n − (µ + 1), and the theorem is true for m = µ + 1. Hence the theorem is true for all m by induction.

SLIDE 36

Proof (continued)

We let H = { u2, . . . , un−µ}. Then

u1 ∈ Span(H ∪ L),

and { u1, . . . , un−µ, v1, . . . , vµ} ⊂ Span(H ∪ L), V = Span( u1, u2, . . . , un−µ, v1, v2, . . . , vµ) ⊂ Span(H ∪ L) ⊂ V , and V = Span(H ∪ L), H ⊂ G, |H| = (n − µ) − 1 = n − (µ + 1), and the theorem is true for m = µ + 1. Hence the theorem is true for all m by induction.