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

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

Step-by-step 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}`