# Prove linearly independent

• Dec 21st 2008, 05:34 PM
ssl000
Prove linearly independent
Prove the statement:

If u, v, and w are linearly independent, then u+v, v, and w are also linearly independent.
• Dec 21st 2008, 05:54 PM
mr fantastic
Quote:

Originally Posted by ssl000
Prove the statement:

If u, v, and w are linearly independent, then u+v, v, and w are also linearly independent.

Given: $\displaystyle \alpha \vec{u} + \beta \vec{v} + \gamma \vec{w} = \vec{0}$ only when $\displaystyle \alpha = \beta = \gamma = 0$ are equal to zero.

Now re-write $\displaystyle \alpha \vec{u} + \beta \vec{v} + \gamma \vec{w}$ as $\displaystyle \alpha (\vec{u} + \vec{v})+ \delta \vec{v} + \gamma \vec{w} = \vec{0}$ ....
• Dec 21st 2008, 06:44 PM
ssl000
Quote:

Originally Posted by mr fantastic
Given: $\displaystyle \alpha \vec{u} + \beta \vec{v} + \gamma \vec{w} = \vec{0}$ where not all of the $\displaystyle \alpha, \beta, \gamma$ are equal to zero.

This can be re-written as $\displaystyle \alpha (\vec{u} + \vec{v})+ \delta \vec{v} + \gamma \vec{w} = \vec{0}$ where $\displaystyle \delta = \beta - \alpha$ ....

Wouldn't that prove that its linearly dependent, since $\displaystyle \delta = \beta - \alpha$?
• Dec 21st 2008, 10:02 PM
mr fantastic
Quote:

Originally Posted by ssl000
Wouldn't that prove that its linearly dependent, since $\displaystyle \delta = \beta - \alpha$?