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

Question

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 $2x_1c+5x_2c=x3_c\Leftrightarrow c(2x_1+5x_2)=cx_3\Leftrightarrow 2x_1+5x_2=x_3.$

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Did I do this correctly?

Answer 1

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.

Answer 2

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

Answer 3

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