Prove linearly independent

**ssl000** · December 21st 2008, 05:34 PM

Prove the statement:

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

**mr fantastic** · December 21st 2008, 05:54 PM

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

Now re-write $\alpha \vec{u} + \beta \vec{v} + \gamma \vec{w}$ as $\alpha (\vec{u} + \vec{v})+ \delta \vec{v} + \gamma \vec{w} = \vec{0}$ ....

**ssl000** · December 21st 2008, 06:44 PM

Originally Posted by mr fantastic

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

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

Wouldn't that prove that its linearly dependent, since $\delta = \beta - \alpha$ ?

**mr fantastic** · December 21st 2008, 10:02 PM

Originally Posted by ssl000

Wouldn't that prove that its linearly dependent, since $\delta = \beta - \alpha$ ?

Sorry I misread the question. I've edited my reply accordingly.

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