Proving two vectors are parallel

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To prove that two vectors ( v ) and ( w ) are parallel if and only if the zero vector is a non-trivial linear combination of ( v ) and ( w ):

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Definitions:

1. Parallel Vectors: Two vectors ( v ) and ( w ) are parallel if there exists a scalar ( k ) such that ( v = k w ) or ( w = k v ).
2. Linear Combination: A vector ( u ) is a linear combination of ( v ) and ( w ) if ( u = \alpha v + \beta w ) for scalars ( \alpha ) and ( \beta ).
3. Non-Trivial Linear Combination: A linear combination is called non-trivial if ( \alpha ) and ( \beta ) are not both zero.
4. Zero Vector: The vector ( \mathbf{0} ) is represented as ( \mathbf{0} = 0 \cdot v + 0 \cdot w ), but for non-trivial combinations, ( \alpha \neq 0 ) or ( \beta \neq 0 ).

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Prove the “If” Direction

Assume the zero vector is a non-trivial linear combination of ( v ) and ( w ):

1. Let ( \alpha v + \beta w = \mathbf{0} ), where ( \alpha \neq 0 ) or ( \beta \neq 0 ).
2. Rearrange the equation:
   [
   \alpha v = -\beta w
   ]
3. Divide both sides by ( \beta ) (assuming ( \beta \neq 0 )) or ( \alpha ) (if ( \beta = 0 )):
   [
   v = -\frac{\beta}{\alpha} w
   ]
   This shows ( v ) is a scalar multiple of ( w ), hence ( v ) and ( w ) are parallel.

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Prove the “Only If” Direction

Assume ( v ) and ( w ) are parallel:

1. If ( v ) and ( w ) are parallel, there exists a scalar ( k ) such that ( v = k w ).
2. Substitute ( v = k w ) into the linear combination ( \alpha v + \beta w ):
   [
   \alpha v + \beta w = \alpha (k w) + \beta w = (\alpha k + \beta) w
   ]
3. To satisfy ( \alpha v + \beta w = \mathbf{0} ), we require ( \alpha k + \beta = 0 ).
   This means ( \beta = -\alpha k ), where ( \alpha \neq 0 ) or ( \beta \neq 0 ).
4. Therefore, the zero vector ( \mathbf{0} ) can be written as:
   [
   \mathbf{0} = \alpha v + \beta w
   ]
   where ( \alpha ) and ( \beta ) are not both zero, proving a non-trivial linear combination.

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Conclusion:

• If the zero vector is a non-trivial linear combination of ( v ) and ( w ), the two vectors are parallel.
• Conversely, if ( v ) and ( w ) are parallel, the zero vector can be expressed as a non-trivial linear combination of ( v ) and ( w ).

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Example:

1. Let ( v = (2, 4) ) and ( w = (1, 2) ).
   Clearly, ( v = 2 w ), so ( v ) and ( w ) are parallel.
2. A non-trivial linear combination yielding the zero vector is:
   [
   \mathbf{0} = 2(-1)(1, 2) + 1(2, 4)
   ]
   Here, ( \alpha = 2 ), ( \beta = -1 ), and ( \alpha, \beta \neq 0 ), confirming the proof.

Proving two vectors are parallel
byu/DigitalSplendid inlearnmath

Understanding Non-Trivial Linear Combinations: How the Zero Vector Reveals Linear Dependence Between Vectors

Disclaimer: This article was created with the assistance of an AI language model and is intended for informational purposes only. Please verify any technical details before implementation.

Understanding the Zero Vector as a Non-Trivial Linear Combination of ( v ) and ( w )

In linear algebra, a linear combination of vectors is a sum of the vectors multiplied by scalars. For example, for vectors ( v ) and ( w ), a linear combination looks like this:

[
\theta v + \beta w
]

Where:
– ( \theta ) and ( \beta ) are scalars (real numbers).
– ( v ) and ( w ) are vectors.

The zero vector, denoted ( \mathbf{0} ), can also be expressed as a linear combination of ( v ) and ( w ). Let’s break down how this works and what it means to be a non-trivial linear combination.

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Trivial vs. Non-Trivial Linear Combinations

1. Trivial Linear Combination:
   When both scalars ( \theta ) and ( \beta ) are 0, the result is always the zero vector:
   [
   \theta v + \beta w = 0v + 0w = \mathbf{0}
   ]
   This is called trivial because it doesn’t provide any meaningful relationship between ( v ) and ( w ).

2. Non-Trivial Linear Combination:
   A non-trivial linear combination is when ( \theta ) and ( \beta ) are not both zero, but the result is still the zero vector:
   [
   \theta v + \beta w = \mathbf{0}, \quad \text{where at least one of } \theta \text{ or } \beta \text{ is not zero}.
   ]
   This implies that ( v ) and ( w ) are related in such a way that they “cancel each other out.” This happens when ( v ) and ( w ) are linearly dependent.

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Step-by-Step Explanation with Example

Let’s consider two vectors ( v ) and ( w ):
[
v = (2, 3), \quad w = (-4, -6)
]

Step 1: Write the linear combination

We want to find scalars ( \theta ) and ( \beta ) such that:
[
\theta v + \beta w = \mathbf{0}
]

Substitute ( v ) and ( w ):
[
\theta (2, 3) + \beta (-4, -6) = (0, 0)
]

Step 2: Solve the equation

This equation can be split into components (x and y):
[
2\theta – 4\beta = 0 \quad \text{(x-component)}
]
[
3\theta – 6\beta = 0 \quad \text{(y-component)}
]

Step 3: Simplify

From the x-component:
[
\theta = 2\beta
]

Substitute ( \theta = 2\beta ) into the y-component:
[
3(2\beta) – 6\beta = 0
]
[
6\beta – 6\beta = 0
]

The equation holds true, meaning ( \theta = 2\beta ) is a valid solution.

Step 4: Non-Trivial Combination

Let ( \beta = 1 ), then ( \theta = 2 ). Substituting these values:
[
2v + 1w = (2, 3) \cdot 2 + (-4, -6) \cdot 1
]
[
= (4, 6) + (-4, -6) = (0, 0)
]

Thus, ( \mathbf{0} = 2v + 1w ), and this is a non-trivial linear combination because ( \theta = 2 ) and ( \beta = 1 ) are not both zero.

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Geometric Interpretation

In this example:
– ( v = (2, 3) ) and ( w = (-4, -6) ) are linearly dependent, meaning one is a scalar multiple of the other:
[
w = -2v
]

When you add ( 2v ) and ( w ), the vectors “cancel out,” leading to the zero vector. This cancellation is what makes the combination non-trivial.

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Key Takeaways

1. A non-trivial linear combination of ( v ) and ( w ) equals the zero vector when ( v ) and ( w ) are linearly dependent.
2. This shows a relationship between ( v ) and ( w ), where one is a scaled version of the other.
3. In practical terms, finding such scalars ( \theta ) and ( \beta ) is a way to test for linear dependence of vectors.

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