Show that $V = \{(x_1, x_1, x_3) \in \mathbb{R}^3 : x_1, x_3 \in \mathbb{R}\}$ is a subspace.
Can someone help with this question? I'm not sure how to approach it.
Show that $V = \{(x_1, x_1, x_3) \in \mathbb{R}^3 : x_1, x_3 \in \mathbb{R}\}$ is a subspace.
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linear-algebra
geometry
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0What is the definition of subspace? – 2017-02-11
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0That the subset is closed under vector addition, scalar multiplication and passes through the 0 vector. I'm just not sure how to start it. – 2017-02-11
2 Answers
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First of all, the zerovector $(0,0,0)$ is in $V$.
Suppose $(x_1,x_1,x_3), (y_1,y_1,y_3) \in V$. Let $\lambda_1, \lambda_2 \in \mathbb{R}$. Then we have the following equalitie: $$\lambda_1(x_1,x_1,x_3) + \lambda_2(y_1,y_1,y_3) = (\lambda_1x_1 + \lambda_2y_1, \lambda_1x_1 + \lambda_2y_1, \lambda_1x_3 + \lambda_2y_3)$$ and this is clearly an element of $V$. Therefore $V$ is a subspace of the vectorspace $\mathbb{R}, \mathbb{R}^3, +$.
As a general answer: whenever you have to check if some set is a subspace, check if the zero element is in the subspace and then take two arbitrary elements and two arbitrary scalars and show that the linear combination is in the given set.
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0"Then we have the following **inequalities**" inequalities? Do you perhaps mean *equality* or *equation* instead? – 2017-02-11
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0Yes yes of course! Thank you! – 2017-02-11
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If you know that the set of solutions of a system of linear homogeneous equations is a subspace (and viceversa), then note that $$ V = \{ (x_1, x_2, x_3) : x_{1} - x_{2} = 0 \}, $$ that is, $V$ is the space of solutions of the one-equation system $$ \begin{cases} x_{1} - x_{2} = 0.\\ \end{cases} $$