I Which of the following sets are linear subspaces of $\mathbb{R}^n$?
- $\{x = (x_1, x_2, \ldots, x_n)| x_1 = a, a = \text{constant}\}$
- $\{x = (x_1, x_2, \ldots, x_n)| x_1 \ge 0\}$
- $\{x = (x_1, x_2, \ldots, x_n)| x_1 \cdot x_2 = 0\}$
- $\{x = (x_1, x_2, \ldots, x_n)| x_1 =0\} \cup \{x = (x_1, x_2, \ldots, x_n)| x_2 =0\} $
- $\{x = (x_1, x_2, \ldots, x_n)| x_1 =0\} \cap \{x = (x_1, x_2, \ldots, x_n)| x_2 =0\} $
My answers and thoughts:
- no, because it has the form $\left(\begin{matrix} 2a \\ z_2 \\ z_3 \end{matrix} \right) \not\in \{x = (x_1, x_2, \ldots, x_n)| x_1 = a, a = \text{constant}\}$
- yes, it's like drawing an $n$-dimensional half-line
- not sure, I think iif $(x_1 = 0) \oplus (x_2 = 0)$
- no, if one of $(x_1,x_2)$ is $0$ in one of the sets, it won't be $0$ in the union
- yes, because $0 \in \mathbb{R}^n$
II Let there be the following subsets of $\mathbb{R}^3$:
- $T_1 = \{ a \in \mathbb{R}^3 | a = (a_1,a_2,a_2), a_1, a_2 \in \mathbb{R}\}$
- $T_2 = \{ a \in \mathbb{R}^3 | a = (a_1,a_2,a_3), a_1 \ge 0\}$
- $T_3 = \{ a \in \mathbb{R}^3 | a = (a_1,a_2,a_3), a_1 + a_2 + a_3 = 0\}$
Which ones are linear subspaces?
My thoughts/answers:
- yes, because given $u, v \in T_1$, it's always true that $ u + v \in T_1$ and $\lambda u \in T_1$
- If I would follow the same train of thought as in I (2) above, I would say yes again, but this time I'm unsure, because it looks like the entire right-half of $\mathbb{R}$, if you imagine it in a cartesian system
- I am inclined to say yes, because they are of the form $\left(\begin{matrix} z_1 \\ z_2 \\ -z_1 - z_2 \end{matrix} \right) \in T_3$, but I'm unsure, as I could express any dimensions in terms of the other two, in which case it's fragments of $\mathbb{R}^3$ whose union is not linear at all
Please let me know which of my answers and thoughts are wrong, what their right answer are and the reasoning behind each. Thanks.