Which of the following sets are subspaces of $\textbf{R}^{3}$:
- A. $\{(x,x+7,x+3) \mid x\in\textbf{R}\}$
- B. $\{(x,y,z) \mid 7x + 3y + 2z = 0\}$
- C. $\{(x,y,z) \mid x,y,z > 0\}$
- D. $\{(x,y,z) \mid −6x − 5y − 4z = 9\}$
- E. $\{(x,0,0) \mid x\in\textbf{R}\}$
- F. $\{(−9x,6x,5x) \mid x\in\textbf{R}\}$
Can someone help me with this question?
I tried solving it with the definition of subspaces, but it didn't work.
I must have done something wrong.
Let's do $B$ as an example. The $0$ vector is there, ok. Now we have to check if it's closed to addition. Let's assume $(x_1,y_1,z_1),(x_2,y_2,z_2)$ are $2$ vectors in that set, i.e their coordinates satisfy $7x+3y+2z=0$. Now what about their sum? The sum of these vectors in $\mathbb{R^3}$ is $(x_1+x_2,y_1+y_2,z_1+z_2)$. And indeed:
$7(x_1+x_2)+3(y_1+y_2)+2(z_1+z_2)=(7x_1+3y_1+2z_1)+(7x_2+3y_2+2y_2)=0+0=0$
So the sum is there. This set is indeed closed to addition.
Now we have to check closure to multiplication by a scalar. Let $(x,y,z)$ be a vector in this set and let $\lambda\in\mathbb{R}$ be any scalar. Then $\lambda(x,y,z)=(\lambda x,\lambda y,\lambda z)$. And we have:
$7(\lambda x)+3(\lambda y)+2(\lambda z)=\lambda(7x+3y+2z)=\lambda\times 0=0$
So $\lambda(x,y,z)$ is in the set as well. So this is indeed a subspace, it is closed under all operations and contains the $0$ vector.