Determine whether $V:=\{x_1,x_2,x_3\in\mathbb{R}^3|2x_1+5x_2=x_3 \}$ is a subspace of $\mathbb{R}^3$

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?

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

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.

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

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

• 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. Commented Jun 18, 2018 at 18:42
• @JMoravitz thanks you I didn’t know Commented Jun 18, 2018 at 18:44
• 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) Commented Jun 18, 2018 at 18:44