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  1. The zero vector is in $V$ since $2\cdot0 + 5\cdot0=0$.
  2. Addition is closed under V. Let $a:=(a_1,a_2,a_3)^T$ and $b:=(b_1,b_2,b_3)^T$. Then we have $2(a+b)_1+5(a+b)_2=2a_1+2b_1+5a_2+5b_2=(2a_1+5a_2)+(2b_1+5b_2)=a_3+b_3=(a+b)_3.$
  3. Multiplication is closed under V. For $c(x_1,x_2,x_3)^T$ we have $2x_1c+5x_2c=x3_c\Leftrightarrow c(2x_1+5x_2)=cx_3\Leftrightarrow 2x_1+5x_2=x_3.$

Did I do this correctly?

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    $\begingroup$ Yes. The notation $(a+b)_1$ is not nice. $\endgroup$ Commented Jun 18, 2018 at 18:35
  • $\begingroup$ @DietrichBurde what would be a better way? Just leave it out? $\endgroup$
    – math_mu
    Commented Jun 18, 2018 at 18:36

3 Answers 3

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Almost. Concerning the third property, it is not true true that$$c(2x_2+5x_2)=cx_3\iff2x_1+5x_2=x_3,$$since $c$ might be $0$. But since all you need to prove is that if $c$ is a scalar, then $(cx_1,cx_2,cx_3)\in V$, that's not a serious problem.

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Consider $V$ as the kernel of $\begin{pmatrix}1&2&-5\end{pmatrix}$.

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It is correct and you can observe that

$V=\{ x\in \mathbb{R}^3: \langle a,x\rangle=0 \}={a}^\perp$

Where $a=(2,5,-1)$ and $\langle ,\rangle$ denotes the standard scalar product on $\mathbb{R}^3$. It is trivially a subspace of $\mathbb{R}^3$ because is the set of the vectors ortogonal with a

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  • $\begingroup$ Typesetting note: :\langle a, x\rangle=0 outputs $:\langle a, x\rangle =0$ and is preferred over using < or > to do the same since the system would insert extra space around these as they are usually used in the context of less than or greater than. $\endgroup$
    – JMoravitz
    Commented Jun 18, 2018 at 18:42
  • $\begingroup$ @JMoravitz thanks you I didn’t know $\endgroup$ Commented Jun 18, 2018 at 18:44
  • $\begingroup$ It would also help to inform the reader of the notation used here for the inner product as it is commonly not encountered by the early parts of a student's first course in linear algebra (though the specific case of the dot product would have been hopefully) $\endgroup$
    – JMoravitz
    Commented Jun 18, 2018 at 18:44

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