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| + | ===SUBSPACE=== | ||
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| + | To be a subspace of vectors the following must be true: | ||
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| + | 1. One set must be a '''subset''' of another set | ||
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| + | 2. The set must be closed under '''scalar multiplication''' | ||
| + | |||
| + | 3. The set must be closed under '''vector addition''' | ||
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| + | |||
== Proving one set is a subset of another set == | == Proving one set is a subset of another set == | ||
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Given sets A and B we say that is is a subset of B if every element of A is also an element of B, that is, | Given sets A and B we say that is is a subset of B if every element of A is also an element of B, that is, | ||
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x∈A implies x∈B | x∈A implies x∈B | ||
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| − | & | + | '''Basic Outline of the Proof that A is a subset of B:''' |
| − | 1. Say | + | |
| + | |||
| + | '''·''' Suppose x ∈ A | ||
| + | |||
| + | 1. Say what it means for x to be in A | ||
| + | |||
2. Mathematical details | 2. Mathematical details | ||
| + | |||
3. Conclude that x satisfies what it means to be in B | 3. Conclude that x satisfies what it means to be in B | ||
| − | & | + | |
| + | |||
| + | '''·''' Conclude x∈B | ||
| + | |||
| + | |||
| + | '''Example''' | ||
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| + | Let A be the set of scalars divisible by 6 and let B be the even numbers. Prove that A is a subset of B. | ||
| + | |||
| + | |||
| + | '''·''' Suppose x ∈ A: | ||
| + | |||
| + | 1. What it means for x to be in A: x = 6k for any scalar k | ||
| + | |||
| + | 2. x = 2 × (3k) | ||
| + | |||
| + | 3k = C | ||
| + | |||
| + | 3. What it means for x to be in B: x = 2C | ||
| + | |||
| + | |||
| + | '''·''' Conclude x∈B | ||
| + | |||
| + | |||
| + | ==Closed Under Scalar Multiplication== | ||
| + | |||
| + | |||
| + | A set of vectors is closed under scalar multiplication if for every '''v'''∈V and every c∈\mathbb{R} we have c'''v'''∈V | ||
| + | |||
| + | |||
| + | '''Basic Outline of the Proof V is Closed Under Scalar Multiplication:''' | ||
| + | |||
| + | |||
| + | |||
| + | '''·''' Suppose '''v'''∈V and c∈\mathbb{R} | ||
| + | |||
| + | 1. Say what it means for '''v''' to be in V | ||
| + | |||
| + | 2. Mathematical details | ||
| + | |||
| + | 3. Conclude that c'''v''' satisfies what it means to be in V | ||
| + | |||
| + | |||
| + | '''·''' Conclude c'''v'''∈V | ||
| + | |||
| + | |||
| + | ==Closed Under Vector Addition== | ||
| + | |||
| + | |||
| + | A set of vectors is closed under vector addition if for every '''v''' and '''w''' ∈ V we have '''v''' + '''w''' ∈ V | ||
| + | |||
| + | |||
| + | '''Basic Outline of the Proof V is Closed Under Vector Addition:''' | ||
| + | |||
| + | |||
| + | '''·''' Suppose '''v''' and '''w''' ∈ V | ||
| + | |||
| + | 1. Say what it means for '''v''' and '''w''' to be in V | ||
| + | |||
| + | 2. Mathematical details | ||
| + | |||
| + | 3. Conclude that '''v'''+ '''w''' satisfies what it means to be in V | ||
| + | |||
| + | |||
| + | '''·''' Conclude '''v''' + '''w''' ∈ V | ||
| + | |||
| + | |||
| + | '''Example''' | ||
| + | |||
| + | Let V be the set of points above the graph of the absolute value function | ||
| + | |||
| + | |||
| + | '''·''' Suppose '''v''' and '''w''' ∈ V | ||
| + | |||
| + | 1. What it means for '''v''' and '''w''' to be in V : | ||
| + | |||
| + | '''v''' = (v1, v2) and v2 > |v1| | ||
| + | |||
| + | '''w''' = (w1, w2) and w2 > |w1| | ||
| + | |||
| + | 2. '''z''' = (z1,z2) = '''v''' + '''w''' = (v1+w1, v2+w2) | ||
| + | |||
| + | So, |z1| = |v1 + w1| ≤ |v1| + |w1| < v2 + w2 = z2 | ||
| + | |||
| + | 3. What it means for '''z''' to be in V: |z1| < z2 | ||
| + | |||
| + | |||
| + | '''·''' Conclude '''z''' = '''v''' + '''w''' ∈ V | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | ---- | ||
| + | ''Explanation of how to determine a subspace. Information referenced from Wabash College MA 223, Spring 2011'' | ||
Latest revision as of 07:48, 25 November 2012
Contents
SUBSPACE
To be a subspace of vectors the following must be true:
1. One set must be a subset of another set
2. The set must be closed under scalar multiplication
3. The set must be closed under vector addition
Proving one set is a subset of another set
Given sets A and B we say that is is a subset of B if every element of A is also an element of B, that is,
x∈A implies x∈B
Basic Outline of the Proof that A is a subset of B:
· Suppose x ∈ A
1. Say what it means for x to be in A
2. Mathematical details
3. Conclude that x satisfies what it means to be in B
· Conclude x∈B
Example
Let A be the set of scalars divisible by 6 and let B be the even numbers. Prove that A is a subset of B.
· Suppose x ∈ A:
1. What it means for x to be in A: x = 6k for any scalar k
2. x = 2 × (3k)
3k = C
3. What it means for x to be in B: x = 2C
· Conclude x∈B
Closed Under Scalar Multiplication
A set of vectors is closed under scalar multiplication if for every v∈V and every c∈\mathbb{R} we have cv∈V
Basic Outline of the Proof V is Closed Under Scalar Multiplication:
· Suppose v∈V and c∈\mathbb{R}
1. Say what it means for v to be in V
2. Mathematical details
3. Conclude that cv satisfies what it means to be in V
· Conclude cv∈V
Closed Under Vector Addition
A set of vectors is closed under vector addition if for every v and w ∈ V we have v + w ∈ V
Basic Outline of the Proof V is Closed Under Vector Addition:
· Suppose v and w ∈ V
1. Say what it means for v and w to be in V
2. Mathematical details
3. Conclude that v+ w satisfies what it means to be in V
· Conclude v + w ∈ V
Example
Let V be the set of points above the graph of the absolute value function
· Suppose v and w ∈ V
1. What it means for v and w to be in V :
v = (v1, v2) and v2 > |v1|
w = (w1, w2) and w2 > |w1|
2. z = (z1,z2) = v + w = (v1+w1, v2+w2)
So, |z1| = |v1 + w1| ≤ |v1| + |w1| < v2 + w2 = z2
3. What it means for z to be in V: |z1| < z2
· Conclude z = v + w ∈ V
Explanation of how to determine a subspace. Information referenced from Wabash College MA 223, Spring 2011
