2
$\begingroup$

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.

  • 0
    Note: I've learned this in German, so perhaps I haven't used the right mathematical terminology everywhere.2012-11-09

1 Answers 1

2

I

  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)

II

  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.

III

  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.

  • 0
    to II 1: the vectors have their y and z coordinates equal, they're both $a_2$2012-11-09
  • 0
    @Flavius Oh, right. I need glasses. Let me edit my answer.2012-11-09
  • 0
    To III: **1.** Do you mean a single point is also a linear subspace? This is insane! (but it sounds extremely consistent with the definition). **2.** lines, surfaces, and I guess you mean that rectangular parallelepipeds are also linear subspaces? What about generally parallelepipeds altogether?2012-11-09
  • 0
    @Flavius Can you be more specific about the point? Because for example, if the point is $(1,1,1)$ then $2 \cdot (1,1,1)$ is already not in the "subspace" (a subspace is also closed with respect to scalar multiplication!).2012-11-09
  • 0
    @Flavius Also: can you be more specific about what you mean by "surface"?2012-11-09
  • 0
    If "linear subspace" means subspace of a vector space, then I.2 and II.2 aren't subspaces.2012-11-09
  • 1
    @JavierBadia Yes, thank you for pointing it out. They aren't closed with respect to scalar multiplication. I'll correct my answer.2012-11-09
  • 0
    @MattN. ignore the point stuff I've said, 1-dimensional is a line, I said a stupidity. By surface I meant a plane. What I don't get now is why the answer to **I 2** is **no**. Do the elements have to always go between $-\infty$ to $\infty$? To **II 2**: no, I can't prove that it's closed with respect to scalar multiplication, so the answer must be **no**. To **II 3**: how do I decide which dimension to express in terms of the others in such cases where an equation is given?2012-11-09
  • 2
    @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.2012-11-09
  • 1
    What Javier said : ) And yes to 1- and 2-dimensional subspaces. Though, the one point space consisting of $0$ is also a linear subspace.2012-11-09