Dimension & Subspace of $\mathbb{R}^3$

[Guest2]

Check whether the following are subspaces of $\mathbb{R}^3$ and if they're find their dimension.

(a) $$x = 0$$, (b) $$x+y = 0$$, (c) $$x+y+z = 0$$, (d) $$x = y$$, (e) $$x = y= z$$, and (f) $$x = y$$ or $$x = z$$.

(a) Let $S = \left\{(x, y, z) \in \mathbb{R}^3:x = 0 \right\}$. I want to check whether $S$ is subspace of $\mathbb{R}^3$. Now, $S$ is nonempty, since $(0,0,0) \in S$. Show that it's closed under addition, we say let $(a_1, a_2, a_3) \in S$ and $(b_1, b_2, b_3) \in S$ then $a_1 = 0$ and $b_1 = 0$ so $a_1+b_1 = 0$ which implies that $(a_1+b_1, a_2+b_2, a_3+b_3) \in S$. To show that it's closed under scalar multiplication, we say let $(a_1, a_2, a_3) \in S$ and $\lambda \in \mathbb{R}$. Then $\lambda a_1 = 0$ therefore $(\lambda a_1, \lambda a_2, \lambda a_3) \in S$. As $S$ is nonempty subset of $\mathbb{R}^3$ that's closed under addition and scalar multiplication, $S$ is a subspace of $\mathbb{R}^3$. Now, how do I find the dimension of $S$? Since we have two free variables, can I say $\text{dim}(S) = 2$?

Or shall I say instead $(x,y,z) = (0,y,z) = (0,y,0)+(0,0,z) = y(0,1,0)+z(0,0,1)$ and since $\left\{(0,1,0), (0,0,1)\right\}$ is linearly independent over $\mathbb{R}^3$, we have $\text{dim}(S) = 2$

Similarly (b),(c),(d),(e) subspaces, though their dimension beats me.

(f) Let $S = \left\{(x, y, z) \in \mathbb{R}^3: x -y= 0 ~\text{or} ~ x-z = 0\right\}$. I want to check whether $S$ is subspace of $\mathbb{R}^3$. Now, $S$ is nonempty, since $(0,0,0) \in S$. Show that it's closed under addition, we say let $(a_1, a_2, a_3) \in S$ and $(b_1, b_2, b_3) \in S$ and then if $x-y=0$ then $a_1 -a_2 =0$ and $b_1-b_2 = 0$ so $(a_1+b_1)-(a_2+b_2) = 0$ which implies that $(a_1+b_1, a_2+b_2, a_3+b_3) \in S$. On the other hand if $x-z = 0$, then $a_1-a_3 = 0$ and $b_1-b_3 = 0$ which gives $(a_1+b_1)-(a_3+b_3) = 0$ which implies $(a_1+b_1, a_2+b_2, a_3+b_3) \in S$ .

To show that it's closed under scalar multiplication, we say let $(a_1, a_2, a_3) \in S$ and $\lambda \in \mathbb{R}$. Then if $x-y = 0$ then $\lambda a_1-\lambda a_2 = 0$ therefore $(\lambda a_1, \lambda a_2, \lambda a_3) \in S$. If on the other hand, $x-z = 0$ then $a_1-a_3 = 0$ and $\lambda a_1-\lambda a_3 = 0$ which implies $(\lambda a_1, \lambda a_2, \lambda a_3) \in S$. As $S$ is nonempty subset of $\mathbb{R}^3$ that's closed under addition and scalar multiplication, $S$ is a subspace of $\mathbb{R}^3$. Now, how do I find the dimension of $S$? Since we have one free variables in each case, can I say $\text{dim}(S) = 1$?

 

[I like Serena]

Hi Guest,

The number of dimensions of a sub space is equal to the number of independent vectors in it.
Alternatively, it's equal to the total dimension minus the dimension of the complement of the sub space.

So for example in (a), we have (0,1,0) and (0,0,1) that are both included in the sub space, while (1,0,0) is excluded from it.
That brings us to a 2-dimensional sub space.

 

[Deveno]

To amplify what I like Serena has posted, the term "free variable" is not usually used as a measure of dimension in Linear Algebra, rather: we need the size (set-cardinality) of any BASIS, which is a maximal linearly independent set, or a MINIMAL spanning set.

Since $\Bbb R^3$ has dimension 3 (one possible basis is $\{(1,0,0),(0,1,0),(0,0,1)\}$, which has 3 elements), obviously the dimension of any subspace is a number between 0 (for the subspace with just the 0-vector in it, $\{(0,0,0)\}$) and 3 (for all of $\Bbb R^3$).

In particular, if we have a non-trivial (more than just the 0-vector) PROPER (not all of $\Bbb R^3$) subspace, our choices are limited: 1 or 2 dimensions (a line through the origin, or a plane through the origin).

 

[Guest2]

Just to be sure that I understand, let me try to find the dimension of (b), (c) and (f).

(b) Dimension of $x+y = 0$ over $\mathbb{R}^3$: we have

$$\begin{aligned} (x, -x, z) & = (x,0,0)+(0,-x,0)+(0,0,z) \\& = x(1,0,0)+x(0,-1,0)+z(0,0,1) \\& = x(1,-1,0)+z(0,0,1) \end{aligned}$$

and $\left\{(1,-1,0),(0,0,1)\right\}$ is linearly independent over $\mathbb{R}^3$, so $\text{dim}(S) = 2.$

(c) Dimension of $x+y+z = 0$ over $\mathbb{R}^3$: we have

$$\begin{aligned} (x, -x-z, z) & = (x,0,0)+(0,-x-z,0)+(0,0,z) \\& = x(1,0,0)+(x+z)(0,-1,0)+z(0,0,1) \\& = x(1,-1,0)+z(0,-1,1) \end{aligned}$$

and $\left\{(1,-1,0),(0,-1,1)\right\}$ is linearly independent over $\mathbb{R}^3$ so $\text{dim}(S) = 2.$

(f) Dimension of $(x-y=0) \cup (x-z = 0)$ over $\mathbb{R}^3$: we have

if $x-y = 0$ we have: $$\begin{aligned} (x, x, z) & = (x,0,0)+(0,x,0)+(0,0,z) \\& = x(1,0,0)+x(0,1,0)+z(0,0,1) \\& = x(1,1,0)+z(0,0,1) \end{aligned}$$

and $\left\{(1,1,0),(0,0,1)\right\}$ is linearly independent over $\mathbb{R}^3$ so $\text{dim}(S) = 2.$

if $x-z = 0$ we have: $$\begin{aligned} (x, y, x) & = (x,0,0)+(0,y,0)+(0,0,x) \\& = x(1,0,0)+y(0,1,0)+x(0,0,1) \\& = x(1,0,1)+y(0,1,0) \end{aligned}$$

and $\left\{(1,0,1),(0,1,0)\right\}$ is linearly independent over $\mathbb{R}^3$, so $\text{dim}(S) = 2.$

 

[Deveno]

Why do you think (f) is a subspace?

Consider:

$(1,-1,0) \in S$ and $(2,0,-2) \in S$, but:

$(1,-1,0) + (2,0,-2) = (3,-1,-2)$-is this in $S$?

 

[Guest2]

Why do you think (f) is a subspace?

Consider:

$(1,-1,0) \in S$ and $(2,0,-2) \in S$, but:

$(1,-1,0) + (2,0,-2) = (3,-1,-2)$-is this in $S$?

But if $(x,y,z) = (1,-1,0)$ then $x-y = 1-(-1) = 2$ and $x-z = 1-0 = 1$.

And if $(x,y,z) = (2,0,-2)$ then $x-y = 2-0 = 2$ and $x-z =2-(-2) = 4$.

(Thinking)

I thought I proved that it was closed under addition as follows:

Let $S = \left\{(x, y, z) \in \mathbb{R}^3: x -y= 0 ~\text{or} ~ x-z = 0\right\}$. I want to check whether $S$ is subspace of $\mathbb{R}^3$. Now, $S$ is nonempty, since $(0,0,0) \in S$. To show that it's closed under addition, I said let $(a_1, a_2, a_3) \in S$ and $(b_1, b_2, b_3) \in S$ and then if $x-y=0$ then $a_1 -a_2 =0$ and $b_1-b_2 = 0$ so $(a_1+b_1)-(a_2+b_2) = 0$ which implies that $(a_1+b_1, a_2+b_2, a_3+b_3) \in S$. On the other hand if $x-z = 0$, then $a_1-a_3 = 0$ and $b_1-b_3 = 0$ which gives $(a_1+b_1)-(a_3+b_3) = 0$ which implies $(a_1+b_1, a_2+b_2, a_3+b_3) \in S$.

But I see that I've missed that the disjunction means one or the other or both. So I didn't consider the both case.

If $x-y=0$ and $x-z = 0$ both, then we have $a_1-a_2 = 0$ and $b_1-b_2 = 0$; and $a_1-a_3 = 0$ and finally $b_1-b_3 = 0.$ This gives $a_1+b_1 = \frac{1}{2}(a_2+a_3+b_2+b_3)$ and also $a_3+b_3 = -\frac{1}{2}(a_2+a_3+b_2+b_3)$ and finally $a_2+b_2 = \frac{1}{2}(a_2+a_3+b_2+b_3)$ and we observe that $(a_1+b_1)-(a_2+b_2) = 0$. But $(a_1+b_1) -(a_3+b_3) =(a_2+a_3+b_2+b_3) \ne 0$ thus $ (a_1+b_1, a_2+b_2, a_3+b_3) \not \in S.$

 

[Deveno]

But if $(x,y,z) = (1,-1,0)$ then $x-y = 1-(-1) = 2$ and $x-z = 1-0 = 1$.

And if $(x,y,z) = (2,0,-2)$ then $x-y = 2-0 = 2$ and $x-z =2-(-2) = 4$.

(Thinking)

I thought I proved that it was closed under addition as follows:

Let $S = \left\{(x, y, z) \in \mathbb{R}^3: x -y= 0 ~\text{or} ~ x-z = 0\right\}$. I want to check whether $S$ is subspace of $\mathbb{R}^3$. Now, $S$ is nonempty, since $(0,0,0) \in S$. To show that it's closed under addition, I said let $(a_1, a_2, a_3) \in S$ and $(b_1, b_2, b_3) \in S$ and then if $x-y=0$ then $a_1 -a_2 =0$ and $b_1-b_2 = 0$ so $(a_1+b_1)-(a_2+b_2) = 0$ which implies that $(a_1+b_1, a_2+b_2, a_3+b_3) \in S$. On the other hand if $x-z = 0$, then $a_1-a_3 = 0$ and $b_1-b_3 = 0$ which gives $(a_1+b_1)-(a_3+b_3) = 0$ which implies $(a_1+b_1, a_2+b_2, a_3+b_3) \in S$.

But I see that I've missed that the disjunction means one or the other or both. So I didn't consider the both case.

If $x-y=0$ and $x-z = 0$ both, then we have $a_1-a_2 = 0$ and $b_1-b_2 = 0$; and $a_1-a_3 = 0$ and finally $b_1-b_3 = 0.$ This gives $a_1+b_1 = \frac{1}{2}(a_2+a_3+b_2+b_3)$ and also $a_3+b_3 = -\frac{1}{2}(a_2+a_3+b_2+b_3)$ and finally $a_2+b_2 = \frac{1}{2}(a_2+a_3+b_2+b_3)$ and we observe that $(a_1+b_1)-(a_2+b_2) = 0$. But $(a_1+b_1) -(a_3+b_3) =(a_2+a_3+b_2+b_3) \ne 0$ thus $ (a_1+b_1, a_2+b_2, a_3+b_3) \not \in S.$

My bad, I mis-read the question, but a similar example still works:

Take $(1,1,0)$ and $(2,0,2)$ Then $(1,1,0) + (2,0,2) = (3,1,2)$ and $x$ is not equal to either $y$ or $z$.

In general, the union of two subspaces is almost never a subspace itself, unless one of them is a subspace of the other one.

The reason being, just because $v_1 \in U$ or $v_1 \in W$ and $v_2 \in U$ or $u_2 \in W$, there is ABSOLUTELY NO REASON to suppose $v_1+v_2$ is in either.

The classic example is this: think of the Euclidean plane $\Bbb R^2$. The $x$-axis, and the $y$-axis are 1-dimensional subspaces. But the union of the two subspaces is just a pair of lines, and if $x,y \neq 0$, the point:

$(x,y) = (x,0) + (0,y)$

is a sum of two axis elements (one on the $x$-axis, and $y$-axis) but it doesn't lie on either one.

So this is one important way in which vector spaces differ from sets without any structure: the smallest set containing both $A$ and $B$ is $A\cup B$, but the smallest vector space containing both $U$ and $W$ is:

$U+W = \{u+w|u \in U, w \in W\}$ which is typically much, much larger than $U \cup W$

 

[Guest2]

Thanks, Deveno.

If $S = \left\{(x,y,z) \in \mathbb{R}^3: x+y+z = 0 ~\text{and}~ x-y-z = 0\right\}$ then $\text{dim}(S) = 2$, right? When checking whether this is closed under addition can we use matrices or perhaps some other tricks? Perhaps I'm working inefficiently but I had a pageful of calculations.

 

[Deveno]

Thanks, Deveno.

If $S = \left\{(x,y,z) \in \mathbb{R}^3: x+y+z = 0 ~\text{and}~ x-y-z = 0\right\}$ then $\text{dim}(S) = 2$, right? When checking whether this is closed under addition can we use matrices or perhaps some other tricks? Perhaps I'm working inefficiently but I had a pageful of calculations.

If $x+y+z = 0$, then $z = -(x+y)$, so all of our vectors of of the form:

$(x,y,-x-y)$.

However, if we simultaneously require (because we said "and", not "or") that:

$x-y-z = 0$, then $z = x - y$ as well, leading to:

$-x-y = x-y$
$-x = x$
$x = 0$.

Thus we are left with:

$y+z = 0 \implies z = -y$
$-y-z = 0 \implies -z = y \implies z = -y$

So our set $S$ is spanned by $(0,1,-1)$ or any non-zero scalar multiple thereof.

Put another way $S = U \cap W$, where $U = \text{span}(\{(1,0,-1),(0,1,-1)\})$ and

$W = \text{span}(\{(1,0,1),(0,1,-1)\}$

Since $(0,1,-1) \in U$ and $W$, we can see that $U \cap W$ contains $\text{span}(\{(0,1,-1)\})$, so it's at least of dimension 1.

Could it be of dimension 2?

Well, each of $U$ and $W$ is of dimension 2, and $U \cap W \subseteq U$ and $U \cap W \subseteq W$.

However, $(1,0,-1) \not\in W$, and $(1,0,1) \not\in U$. So $U \cap W$ is strictly smaller than either $U$ or $W$, and thus has dimension < 2.

(You may want to take this opportunity to convince yourself that if $S$ is a vector space of dimension $k$, and $T$ is a subspace of $S$ of dimension $k$, then $T = S$, so if $T$ is a subspace of $S$ and $T \neq S$, it follows $T$ has dimension $< k$).