Question

0 0 Determine whether the set O 0 is a basis for R3. If the set is not a basis, determine whether the set is linearly independent and whether the set spans R3. 0 Which of the following describe the set? Select all that apply. A. The set is a basis for R3. B. The set is linearly independent. C. The set spans R3. D. None of the above are true.

Answer 1

Definitions:

Let be a vector space over the field $\mathbb{F}$ and
{V1, V2, ..., Un} CV
.

1). Then we say that the set $\{v_1,v_2,...,v_n\}$ is Linearly Independent if

  $c_1v_1+c_2v_2+...+c_nv_n=0\ \Rightarrow c_1=c_2=...=c_n=0$ where

  $each\ c_i\in \mathbb{F},i=1,2,...,n$.

2). We say the set $\{v_1,v_2,...,v_n\}$ spans if for any $v\in V$ there exists $c_1,c_2,...,c_n\in \mathbb{F}$

such that  $v=c_1v_1+c_2v_2+...+c_nv_n$.

3). we say the set $\{v_1,v_2,...,v_n\}$ is a basis for   if $\{v_1,v_2,...,v_n\}$ is linearly independent and

  $\{v_1,v_2,...,v_n\}$ spans  .

Given that $\begin{Bmatrix} \begin{bmatrix} 0\\0 \\1 \end{bmatrix},\begin{bmatrix} 1\\0 \\1 \end{bmatrix},\begin{bmatrix} 0\\0 \\0 \end{bmatrix} \end{Bmatrix}\subset \mathbb{R}^3$ .

Then $\begin{Bmatrix} \begin{bmatrix} 0\\0 \\1 \end{bmatrix},\begin{bmatrix} 1\\0 \\1 \end{bmatrix},\begin{bmatrix} 0\\0 \\0 \end{bmatrix} \end{Bmatrix}$ is not a basis.

The set  $\begin{Bmatrix} \begin{bmatrix} 0\\0 \\1 \end{bmatrix},\begin{bmatrix} 1\\0 \\1 \end{bmatrix},\begin{bmatrix} 0\\0 \\0 \end{bmatrix} \end{Bmatrix}$ is not Linearly Independent because for any scalar  $c(\neq 0)\in \mathbb{R}$ we have

$0 \begin{bmatrix} 0\\0 \\1 \end{bmatrix}+0\begin{bmatrix} 1\\0 \\1 \end{bmatrix}+c\begin{bmatrix} 0\\0 \\0 \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$ which violating definition (1).

Also the set $\begin{Bmatrix} \begin{bmatrix} 0\\0 \\1 \end{bmatrix},\begin{bmatrix} 1\\0 \\1 \end{bmatrix},\begin{bmatrix} 0\\0 \\0 \end{bmatrix} \end{Bmatrix}$ does not spans because for any vector $\begin{bmatrix} 0\\a \\0 \end{bmatrix}\in \mathbb{R}^3 \ where\ a(\neq 0)\in \mathbb{R}$

there does not exists scalars $c_1,c_2,c_3\in \mathbb{R}$ such that $c_1\begin{bmatrix} 0\\0 \\1 \end{bmatrix}+c_2\begin{bmatrix} 1\\0 \\1 \end{bmatrix}+c_3\begin{bmatrix} 0\\0 \\0 \end{bmatrix}=\begin{bmatrix} 0\\a \\0 \end{bmatrix}$ .

As is there exists such scalars then

  $c_1\begin{bmatrix} 0\\0 \\1 \end{bmatrix}+c_2\begin{bmatrix} 1\\0 \\1 \end{bmatrix}+c_3\begin{bmatrix} 0\\0 \\0 \end{bmatrix}=\begin{bmatrix} 0\\a \\0 \end{bmatrix}$

$\Rightarrow \left.\begin{matrix} c_2=0\\0=a\\c_1+c_2=0 \end{matrix}\right\}$

which is a contradiction as $a\neq0$ .

Hence option (D) is correct.