# Subsets in Linear Algebra

Looking to get some help with some problems. Not quite understanding the concepts of vectors subspaces. So I need to determine which of the following are subsets in $\Bbb R^3$ are actually subspaces.

1) $U = \{(x, y, z) \in \Bbb R^3 : x=y\}$
2) $U = \{(x, y, z) \in \Bbb R^3 : xyz = 0 \}$
3) $U = \{(x, y, z) \in \Bbb R^3 : x+ y + z = 0 \}$
4) $U = \{(x, y, z) \in \Bbb R^3 : x \leq y \leq z \}$

I know that

$u + v \in V$
$u + v = v + u$
$(u + v )= u + v$
$(u+v) + w = u+ (v + w)$
$0 + u = u$
$a u \in V$
$a(bu) = (ab)u$
$1 u = u$
$a(u+v) = au + av$
$(-u)+ u = 0$

• any idea on solving thisproblem? Commented Sep 27, 2016 at 19:54
• When I see a problem like this, I try to visualize the set. If it's not a flat, linear space, then I try to look for two elements that will violate closure under addition. Commented Sep 27, 2016 at 20:31

I'll show you 1), if you get stuck with the others, let me know.

$U$ consists of all the vectors for which the first coordinate is equal to the second coordinate. We have to check that for all $u,v \in U$ it is true that:

• $U$ is not empty
• $u+v \in U$
• for all $a \in \mathbb R$, $au \in U$

Let's get started.

• $(1,1,0) \in U$, so $U$ is not empty
• $(x_1, y_1, z_1) + (x_2, y_2, z_2) = (x_1+x_2, y_1+y_2, z_1+z_2)$ so if $x_1 = y_1$ and $x_2 = y_2$, then $x_1 + x_2 = y_1 + y_2$, therefore the second condition is also statisfied.
• if $u= (x,y,z)\in U$, then $au = (ax,ay,az) \in U$, thus the third condition is also met.

Edit: Above I'm using the following definition.

Let $V$ be a vector space over the field of let's say the real numbers, e.g. $V= \mathbb R^3$. A subset $U \subseteq V$ is a subspace of $V$ if and only if the following three conditions are met

• $U$ is not empty
• $U$ is closed under addition
• $U$ is closed under multiplication by scalars (i.e. real numbers)
• are the list of conditions I listed above not correct for this question? I am struggling with the other 3 but trying to comprehend your explanation. Commented Sep 27, 2016 at 20:02
• for number 2 xyz = 0 so one of x y z must =0 so howdoes this effect the problem Commented Sep 27, 2016 at 20:09
• I've added the definition of a subspace in a vector space. Hope that helps. What you've listed are properties of the vector space $V$, you don't have to check all those when you want to decide if $U$ is a subspace of $V$ or not Commented Sep 27, 2016 at 20:10
• okay thank you for your time ill try the next couple Commented Sep 27, 2016 at 20:12
• Commented Sep 27, 2016 at 20:18

Since some basic axioms of a vector space, such as "u+ v= v+ u", "(u+ v)+ w= u+ (v+ w)", "u+ 0= u", etc. are true for all vectors in the space they are necessarily true for vectors in the sub-set. To show that a sub-set of a vector space is a sub-space, you only need to prove it is "closed" under vector addition and scalar multiplication- that is the sum of two vectors in the sub-set is again in the subset and the product of a scalar and a vector in the sub-set is in the sub-set, and that the set is non-empty (showing that it contains the 0 vector will do that).

1) $U= \{(x,y,z)∈R^3:x=y\}$ The 0 vector is (0, 0, 0) and 0= 0. If (a, b, c) and (x, y, z) are two vectors in this set then a= b and x= y. Then their sum is (a+ x, b+ y, c+ z). Since a= b, a+ x= b+ x. Since x= y, b+ x= b+ y so a+ x= b+ y. If p is any number then p(x, y, z)= (px, py, pz). Since x= y, px= py.

2) $U=\{(x,y,z)∈R^3:xyz=0\}$ The 0 vector is (0, 0, 0) and 0(0)(0)= 0. However, (1, 0, 1) and (0, 1, 0) are in this set while their sum, (1, 1, 1) is not. This is NOT a subspace.

3) $U=\{(x,y,z)∈ R^3:x+y+z=0\}$ The 0 vector, (0, 0, 0) satisfies 0+ 0+ 0= 0 so is in the set. If (a, b, c) and (x, y, z) are such that a+ b+ c= 0 and x+ y+ z= 0 then the sum of the two vectors is (a+ x, b+ y, c+ z) and (a+ x)+ (b+ y)+ (c+ z)= (a+ b+ c)+ (x+ y+ z)= 0+ 0= 0. And if p is any real number, p(x, y, z)= (px, py, pz) and px+ py+ pz= p(x+ y+ z)= p(0)= 0.

4) $U=\{(x,y,z)∈ R^3:x≤y≤z\}$ The 0 vector, (0, 0, 0)) satisfies $0\le 0\le 0$ so is in the set. If (a, b, c) and (x, y, z) are such that $a\le b\le c$ and $x\le y\le z$. then the sum of the two vectors is (a+ x, b+ y, c+ z). Now, since $a\le b$, $a+ x\le b+ x$. Further since $x\le y$, $x+ b\le y+ b$. Putting those together, $a+ x\le y+ b$. Now, since $b\le c$, $b+ y\le c+ y$. Since $y\le z$, $y+ c\le z+ c$ so $b+ y\le z+ c$. Together, $a+ x\le b+ y\le z+ c$. [b]However[/b] if p= -1 and (x, y, z)= (1, 2, 3), then p(x, y, z)= -1(1, 2, 3)= (-1, -2, -3). But -1> -2> -3 so this is NOT a subspace.

• thanks for your answer and time Commented Sep 27, 2016 at 20:58

The main idea of a vector subspace is that it is a subset of the set of vectors that happens to be a vector space over the same field with the same binary operations established in the "bigger" vector space. So for a subset of a vector space be a vector space, what should happen is closure under vector addition (that is, given any two vector spaces, their sum must be in the considered set), closure under scalar multiplication (that is: given a vector from the considered subset and a scalar from the field over the "bigger" set is vector space, their multiplication should be in the bigger space), also there must exist an identity elements for the sum, and since the operation considered is the same as the original factor space and the identity vector for sum is a unique, it follows that $0$ must be in the subset. Observe that the other properties that vector spaces should satisfy follow from the properties of the defined operation on the original vector space. Summarizing, a subset of a vector space is a subspace if it is closed under addition and scalar multiplication and the identity for the sum is in it.