Question

The following statement is either true​ (in all​ cases) or false​ (for at least one​ example). If​ false, construct a specific example to show that the statement is not always true. Such an example is called a counterexample to the statement. If a statement is​ true, give a justification. If v1 and v2 are in set of real numbers R Superscript 4 and v2 is not a scalar multiple of v1​, then ​{v1​,v2​} is linearly independent. Choose the correct answer below. A. The statement is false. The vector v1 could be a scalar multiple of vector v2. B. The statement is false. The vector v1 could be the zero vector. C. The statement is true. A set of vectors is linearly independent if and only if none of the vectors are a scalar multiple of another vector. D. The statement is false. The vector v1 could be equal to the vector v2.

Answer 1

Answer:

• B. The statement is false. The vector v1 could be the zero vector.

Explanation:

Two vectors $\vec{v}_1$ and $\vec{v}_2$ are linearly independent if the equation

$\alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 = \vec{0}$

where $\alpha_1$ and $\alpha_2$ are scalars, has only one solution:

$\alpha_1 = \alpha_2 = 0$.

If there is more than one solution, we say that the vector are linearly dependent.

Why the answer is B. :

if $\vec{v}_1 = 0$ then, $\alpha_1$ could have any value of the scalar group, as for any scalar $\alpha$

$\alpha * \vec{0} = \vec{0}$

So, we get that there is more than one solution.

So, a particular counterexample is:

$\vec{v}_1  = (0,0,0,0)$

$\vec{v}_2 = (1,0,0,0)$

as $\vec{v}_2$ is not an scalar multiple of $\vec{v}_2$, and the equation

$\alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 = \vec{0}$

has as solution

$\alpha_1 = 2$

$\alpha_2 = 0$

as we can see

$2 (0,0,0,0) + 0 (1,0,0,0) = \vec{0}$

$(2* 0,2 *0,2*0,2*0) +  (0*1,0*0,0*0,0*0) = \vec{0}$

$(0,0,0,0) +  (0,0,0,0) = \vec{0}$

$(0+0,0+0,0+0,0+0) = \vec{0}$