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I Which of the following sets are linear subspaces of $\mathbb{R}^n$?

  1. $\{x = (x_1, x_2, \ldots, x_n)| x_1 = a, a = \text{constant}\}$
  2. $\{x = (x_1, x_2, \ldots, x_n)| x_1 \ge 0\}$
  3. $\{x = (x_1, x_2, \ldots, x_n)| x_1 \cdot x_2 = 0\}$
  4. $\{x = (x_1, x_2, \ldots, x_n)| x_1 =0\} \cup \{x = (x_1, x_2, \ldots, x_n)| x_2 =0\} $
  5. $\{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:

  1. no, because it has the form $\left(\begin{matrix} 2a \\ z_2 \\ z_3 \end{matrix} \right) \not\in \{x = (x_1, x_2, \ldots, x_n)| x_1 = a, a = \text{constant}\}$
  2. yes, it's like drawing an $n$-dimensional half-line
  3. not sure, I think iif $(x_1 = 0) \oplus (x_2 = 0)$
  4. no, if one of $(x_1,x_2)$ is $0$ in one of the sets, it won't be $0$ in the union
  5. yes, because $0 \in \mathbb{R}^n$

II Let there be the following subsets of $\mathbb{R}^3$:

  1. $T_1 = \{ a \in \mathbb{R}^3 | a = (a_1,a_2,a_2), a_1, a_2 \in \mathbb{R}\}$
  2. $T_2 = \{ a \in \mathbb{R}^3 | a = (a_1,a_2,a_3), a_1 \ge 0\}$
  3. $T_3 = \{ a \in \mathbb{R}^3 | a = (a_1,a_2,a_3), a_1 + a_2 + a_3 = 0\}$

Which ones are linear subspaces?

My thoughts/answers:

  1. yes, because given $u, v \in T_1$, it's always true that $ u + v \in T_1$ and $\lambda u \in T_1$
  2. 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
  3. 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.

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Note: I've learned this in German, so perhaps I haven't used the right mathematical terminology everywhere. – Flavius Nov 9 '12 at 18:33
up vote 2 down vote accepted


  1. correct

  2. not correct

  3. yes: since $x_1 x_2 = 0$ if either $x_1 = 0$ or $x_2 = 0$ this is the same as 4.

  4. yes: this is just the whole space.

  5. yes: this is the $n-2$-dimensional subspace (since the first $2$ coordinates are zero)


  1. correct

  2. You can prove that it's closed with respect to $+$ and contains $0$ as well as inverses. Can you also prove that it's closed with respect to scalar multiplication?

  3. yes! Draw a picture.


  1. What do linear subspaces in $\mathbb R^3$ look like? For example, what do $1$-dimensional subspaces look like? And what do $2$-dimensional ones look like?

Hope this helps.

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to II 1: the vectors have their y and z coordinates equal, they're both $a_2$ – Flavius Nov 9 '12 at 19:06
@Flavius Oh, right. I need glasses. Let me edit my answer. – Rudy the Reindeer Nov 9 '12 at 19:07
@JavierBadia Yes, thank you for pointing it out. They aren't closed with respect to scalar multiplication. I'll correct my answer. – Rudy the Reindeer Nov 9 '12 at 20:09
@Flavius: I think you are, if you will, thinking too much. You have a perfectly good defintion: Given a vector space $V$, a set $S \subset V$ is a subspace if $0 \in S$ and S is closed with respect to addition and scalar multiplication. Use that. – Javier Nov 9 '12 at 21:03
What Javier said : ) And yes to 1- and 2-dimensional subspaces. Though, the one point space consisting of $0$ is also a linear subspace. – Rudy the Reindeer Nov 9 '12 at 21:15

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