Question

If the subset $\left\{u,v,w\right\}$of V is linearly independent over F, prove that $\mathbf{\{}\mathbf{u}\mathbf{,}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{,}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{+}\mathbf{w}\mathbf{\}}$is linearly independent.

Short answer

Answer:

$\mathbf{\{}\mathbf{u}\mathbf{,}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{,}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{+}\mathbf{w}\mathbf{\}}$is linearly independent.

Step-by-step solution

Definition of linearly independent.

A set of vector is said to be linearly dependent if one of the vector of the set can be written as a linear combination of the others, and if no vector can be written as a linear combination of others then the vector is said to be linearly independent.

Showing that {u,u+v,u+v+w} is linearly independent.

The set $\left\{u,v,w\right\}$ is given to be linearly independent over F.

As $\mathbf{\{}\mathbf{u}\mathbf{,}\mathbf{v}\mathbf{,}\mathbf{w}\mathbf{\}}$is linearly independent. So,

For $\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}\mathbf{\in }\mathit{F}$

$\mathit{a}\mathit{u}\mathbf{+}\mathit{b}\mathit{v}\mathbf{+}\mathit{c}\mathit{w}\mathbf{=}\mathbf{0}$

Such that $\mathit{a}\mathbf{=}\mathit{b}\mathbf{=}\mathit{c}\mathbf{=}\mathbf{0}$

Now the consider the set $\mathbf{\{}\mathbf{u}\mathbf{,}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{,}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{+}\mathbf{w}\mathbf{\}}$

For this set to be linearly independent, the coefficients of its element should be zero. Therefore,

${\mathit{C}}_{\mathbf{1}}\mathit{u}\mathbf{+}{\mathit{C}}_{\mathbf{2}}\mathbf{(}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{)}\mathbf{+}{\mathit{C}}_{\mathbf{3}}\mathbf{(}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{+}\mathbf{w}\mathbf{)}\mathbf{=}\mathbf{(}{\mathbf{C}}_{\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}\mathbf{+}{\mathbf{C}}_{\mathbf{3}}\mathbf{)}\mathit{u}\mathbf{+}\mathbf{(}{\mathbf{C}}_{\mathbf{2}}\mathbf{+}{\mathbf{C}}_{\mathbf{3}}\mathbf{)}\mathit{v}\mathbf{+}{\mathit{C}}_{\mathbf{3}}\mathit{w}$

As $\mathbf{\{}\mathbf{u}\mathbf{,}\mathbf{v}\mathbf{,}\mathbf{w}\mathbf{\}}$is linearly independent, the coefficient of its elelmentshoul be zero. Therefore,

$$\begin{gathered} {\mathit{C}}_{\mathbf{1}}\mathbf{+}{\mathit{C}}_{\mathbf{2}}\mathbf{+}{\mathit{C}}_{\mathbf{3}}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{C}}_{\mathbf{2}}\mathbf{+}{\mathit{C}}_{\mathbf{3}}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{C}}_{\mathbf{3}}\mathbf{=}\mathbf{0} \end{gathered}$$

This implies

${\mathit{C}}_{\mathbf{1}}\mathbf{=}{\mathit{C}}_{\mathbf{2}}\mathbf{=}{\mathit{C}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$

Thus the coefficients of its elements are zero. Hence $\mathbf{\{}\mathbf{u}\mathbf{,}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{,}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{+}\mathbf{w}\mathbf{\}}$is linearly independent.