The given operation is not a vector space because it fails axiom 8 ("distributivity of scalar multiplication with respect to field addition").
<h3>What is a vector space?</h3>
A vector space can be defined as a space (set) which comprises vectors, whose elements can be added under the associative and commutative operation, and can be multiplied by scalars under the associative and distributive operation.
This ultimately implies that, a vector space must be comprised of at least one element, which is generally regarded as its zero vector and the smallest possible vector space.
For every element {u, v and w} in vector (V), and element {a and b} in vector (F), this 8th axiom must be satisfied in order to have a vector space:
(k + m)u = ku + mu
<u>Note:</u> The above axiom (k + m)u = ku + mu is generally referred to as the "distributivity of scalar multiplication with respect to field addition."
<u>Given the following vector:</u>
k(x, y, z) = {k²x, k²y, k²z}
If x₁ · x₂ ≥ 0, then, (kx₁) · (kx₁) = k²x₁y₁ ≥ 0 [Closed in scalar multiplication].
If x₁ · x₂ ≥ 0, x₂ · y₂ ≥ 0, then (x₁ + x₂) · (y₁ + y₂):
x₁y₁ + x₂y₂ + x₁y₂ + x₂y₁ < 0
x₁y₁ + x₂y₂ < -(x₁y₂ + x₂y₁) [Not closed in vector addition].
In conclusion, we can infer and logically deduce that the given operation is not a vector space because it fails axiom 8 ("distributivity of scalar multiplication with respect to field addition").
Read more on vector space here: brainly.com/question/11383
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Complete Question:
Determine whether each set equipped with the given operations is a vector space. For those that are not vector spaces identify the vector space axioms that fail.
The set of all triples of real numbers with the standard vector addition but with scalar multiplication defined by k(x, y, z) = {k²x, k²y, k²z}.