- Thread starter
- #1

- Apr 14, 2013

- 4,114

We have that the vectrs $\vec{v},\vec{w}, \vec{u}$ are linearly independent.

I want to check if the pairs

- $\vec{v}, \vec{v}+\vec{w}$
- $\vec{v}+\vec{u}$, $\vec{w}+\vec{u}$
- $\vec{v}+\vec{w}$, $\vec{v}-\vec{w}$

Since $\vec{v}, \vec{w}, \vec{u}$ are linearly independet it holds that $\lambda_1\vec{v}+\lambda_2\vec{w}+\lambda_3\vec{u}=0 \Rightarrow \lambda_1=\lambda_2=\lambda_3=0$ ($\star$).

We have the following:

- $\vec{v}, \vec{v}+\vec{w}$ :

$\alpha_1\vec{v}+\alpha_2(\vec{v}+\vec{w})=0 \Rightarrow (\alpha_1+\alpha_2)\vec{v}+\alpha_2\vec{w}=0$

How can we continue here?

$$$$

- $\vec{v}+\vec{u}$, $\vec{w}+\vec{u}$ :

$\alpha_1(\vec{v}+\vec{u})+\alpha_2(\vec{w}+\vec{u})=0 \Rightarrow \alpha_1\vec{v}+(\alpha_1+\alpha_2)\vec{u}+\alpha_2\vec{w}=0$

From ($\star$) it folows that $\alpha_1=\alpha_1+\alpha_2=\alpha_2=0\Rightarrow \alpha_1=\alpha_2=0$ and so this means that the vectors $\vec{v}+\vec{u}$ and $\vec{w}+\vec{u}$ are linearly independent.

$$$$

- $\vec{v}+\vec{w}$, $\vec{v}-\vec{w}$ :

$\alpha_1(\vec{v}+\vec{w})+\alpha_2(\vec{v}-\vec{w})=0\Rightarrow (\alpha_1+\alpha_2)\vec{v}+(\alpha_1-\alpha_2)\vec{w}=0$

How can we continue here?