Finding which sets are subspaces of R3

Question

Hello. I have attached an image of the question I am having trouble with. I thought that it was 1,2 and 6 that were subspaces of $\mathbb R^3$.

Here is my working:

1. Rearranged equation ---> $x+y-z=0$. Is a subspace since it is the set of solutions to a homogeneous linear equation.

2. $0$ is in the set if $x=y=0$. Is a subspace. (I know that to be a subspace, it must be closed under scalar multiplication and vector addition, but there was no equation linking the variables, so I just jumped into thinking it would be a subspace...)

3. Rearranged equation ---> $xy - xz=0$. $0$ is in the set if $x=0$ and $y=z$. I said that $(1,2,3)$ element of $R^3$ since $x,y,z$ are all real numbers, but when putting this into the rearranged equation, there was a contradiction. So, not a subspace.

4. = space $\{\,(1,0,0),(0,0,1)\,\}$. set is not a subspace (no zero vector)

5. Similar to above.

6. $0$ is in the set if $m=0$. Again, I was not sure how to check if it is closed under vector addition and multiplication. From seeing that $0$ is in the set, I claimed it was a subspace.

I appreciate any help. Thank you.

Answer 1

Guide:

For part $4$, note that

$U_4=\operatorname{Span}\{ (1,0,0), (0,0,1)\}$, it is written in the form of span of elements of $\mathbb{R}^3$ which is closed under addition and scalar multiplication. Hence it is a subspace. I will leave part $5$ as an exercise.

Alternatively, let me prove $U_4$ is a subspace by verifying it is closed under additon and scalar multiplicaiton explicitly.

Closed under addition: Let $x \in U_4$, $\exists s_x, t_x$ such that $x=s_x(1,0,0)+t_x(0,0,1)$ . Let $y \in U_4$, $\exists s_y, t_y$ such that $y=s_y(1,0,0)+t_y(0,0,1)$, then $x+y = (s_x+s_y)(1,0,0)+(s_y+t_y)(0,0,1)$ but we have $s_x+s_y, t_x+t_y \in \mathbb{R}$, hence $x+y \in U_4$.

Closed under scalar multiplication, let $c \in \mathbb{R}$, $cx = (cs_x)(1,0,0)+(ct_x)(0,0,1)$ but we have $cs_x, ct_x \in \mathbb{R}$, hence $cx \in U_4$.

Checking whether the zero vector is in is not sufficient.

Try to exhibit counter examples for part $2,3,6$ to prove that they are either not closed under addition or scalar multiplication. For example, for part $2$, $(1,1,1) \in U_2$, what about $\frac12 (1,1,1)$, is it in $U_2$?

Answer 2

Here are the definitions I think you are missing:

Closure under vector addition:

A subset $S$ of $\mathbb{R}^3$ is closed under vector addition if the sum of any two vectors in $S$ is also in $S$. In other words, if $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are in the subspace, then so is $(x_1+x_2,y_1+y_2,z_1+z_2)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $x_1,y_1,x_2,y_2\in\mathbb{R}$, the vector $(x_1,y_2,x_1y_1)+(x_2,y_2,x_2y_2)=(x_1+x_2,y_1+y_2,x_1x_2+y_1y_2)$ is in the subset.

Closure under scalar multiplication:

A subset $S$ of $\mathbb{R}^3$ is closed under scalar multiplication if any real multiple of any vector in $S$ is also in $S$. In other words, if $r$ is any real number and $(x_1,y_1,z_1)$ is in the subspace, then so is $(rx_1,ry_1,rz_1)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $r,x_1,y_1\in\mathbb{R}$, the vector $(rx_1,ry_2,rx_1y_1)$ is in the subset.

The set $\{s(1,0,0)+t(0,0,1)|s,t\in\mathbb{R}\}$ from problem 4 is the set of vectors that can be expressed in the form $s(1,0,0)+t(0,0,1)$ for some pair of real numbers $s,t\in\mathbb{R}$. A similar definition holds for problem 5.

(Also I don't follow your reasoning at all for 3.)