Question

Given the set: W={(𝑥,𝑦,𝑧) ∈ 𝐼𝑅^3 / 2𝑥 − 𝑦 =0} a) Prove that the set is a vector subspace in IR3 b) Determine the generator<W>

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Answer to a math question Given the set: W={(𝑥,𝑦,𝑧) ∈ 𝐼𝑅^3 / 2𝑥 − 𝑦 =0} a) Prove that the set is a vector subspace in IR3 b) Determine the generator<W>

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Jett
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96 Answers
a) Para demostrar que W es un subespacio vectorial en \mathbb{R}^3, debemos verificar que cumple con las siguientes propiedades:

1. El vector cero pertenece a W: Como 2(0) - 0 = 0, el vector cero pertenece a W.

2. Cerrado bajo la suma: Sean (x_1, y_1, z_1) y (x_2, y_2, z_2) dos elementos arbitrarios en W, entonces:
2x_1 - y_1 = 0
2x_2 - y_2 = 0
Ahora sumamos estos dos elementos:
2(x_1+x_2) - (y_1+y_2) = 2x_1 - y_1 + 2x_2 - y_2 = 0 + 0 = 0
Por lo tanto, la suma está también en W.

3. Cerrado bajo la multiplicación por un escalar: Sea (x, y, z) \in W y \alpha un escalar arbitrario, entonces:
2x - y = 0
Ahora multiplicamos por \alpha:
2(\alpha x) - (\alpha y) = \alpha(2x - y) = \alpha(0) = 0
Por lo tanto, la multiplicación está también en W.

Dado que W cumple con las tres propiedades, podemos concluir que W es un subespacio vectorial en \mathbb{R}^3.

b) Para determinar el generador del conjunto W, necesitamos encontrar el conjunto más pequeño que genera W. Dado que W está definido por la ecuación 2x - y = 0, podemos expresar y como función de x:
2x - y = 0 \Rightarrow y = 2x
Por lo tanto, cualquier vector en W se puede representar como (x, 2x, z) con x, z \in \mathbb{R}.

Finalmente, el generador de W está dado por:
\langle W \rangle = \{(x, 2x, z) \ | \ x, z \in \mathbb{R}\}

\boxed{\langle W \rangle = \{(x, 2x, z) \ | \ x, z \in \mathbb{R}\}}

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