# Subspaces (basic)

• April 4th 2009, 04:49 AM
U-God
Subspaces (basic)
I have to find which of these subsets are subspaces of $R^3$

$a)$ $(a,2a,3a) \mid a \epsilon R$
$b)$ $(x,y,z) \mid 2x - 3y + 4z = 1$
$c)$ $(1,a,b) \mid a,b \epsilon R$
$d)$ $(a-2b,b,a+b) \mid a,b \epsilon R$
$e)$ $(x,y,z) \mid 2x - z = 0$

I know that for a subset to be a subspace of a given set, it can't be an empty set, it must be closed under addition and closed under scalar multiplication.
I can see that $a)$ is going to be a subspace of $R^3$ because if $a$ is real, then a real number plus a real number will yield a real number. Also, a real number multiplied by a real number will yield a real number.

But I don't know how to prove it, nor how to analyse any of the others algebraically.

• April 4th 2009, 05:31 AM
scorpion007
For the first one: $S=\lbrace (a,2a,3a) \mid a \in R \rbrace \subset R^3$.

Closure under vector addition:

Let $\vec{x}=(a_1,2a_1,3a_1), ~\vec{y}= (a_2,2a_2,3a_2)\in S$.
$\vec{x}+\vec{y}=(a_1+a_2, 2a_1+2a_2, 3a_1+3a_2)$
Grouping coefficients,
$= (a_1+a_2,2(a_1+a_2), 3(a_1+a_2)) \in S$.
Thus S is closed under vector addition.

Scalar multiplication can be showed similarly. (Hint, define an $\alpha \in R$ and show that $\forall \vec{x} \in S,~ \alpha \vec{x} \in S$).
• April 4th 2009, 08:02 AM
Twig
You should also state that the zero vector of R^3 is in your subspace.
This is usually one of the first checks on makes if a subset is a subspace.
Because if the zero vector is not there, then you don´t even have to test for vector addition and scalar multiplication.
• April 5th 2009, 04:22 AM
U-God
Quote:

Originally Posted by scorpion007
For the first one: $S=\lbrace (a,2a,3a) \mid a \in R \rbrace \subset R^3$.

Closure under vector addition:

Let $\vec{x}=(a_1,2a_1,3a_1), ~\vec{y}= (a_2,2a_2,3a_2)\in S$.
$\vec{x}+\vec{y}=(a_1+a_2, 2a_1+2a_2, 3a_1+3a_2)$
Grouping coefficients,
$= (a_1+a_2,2(a_1+a_2), 3(a_1+a_2)) \in S$.
Thus S is closed under vector addition.

Scalar multiplication can be showed similarly. (Hint, define an $\alpha \in R$ and show that $\forall \vec{x} \in S,~ \alpha \vec{x} \in S$).

So if $\vec{x}=(a_1,2a_1,3a_1)$still,
I can say that $\alpha \vec{x} = \alpha (a_1,2a_1,3a_1)$
therefore $\alpha \vec{x} = (\alpha a_1, 2 \alpha a_1, 3 \alpha a_1)$

Is that proven though?
I didn't understand what you meant with the hint sorry.
Thanks,
• April 5th 2009, 04:42 AM
scorpion007
Indeed, $\alpha \vec{x} = (\alpha a, 2 (\alpha a), 3 (\alpha a))
$
. But, we know that any vector x in S has its first component: any real, the second component: twice the first component, and the third component: three times the first. In other words, any vector $\vec{v} = (a, 2a, 3a) \in S$ because it satisfies the conditions of membership of S.

We can see that $\alpha \vec{x}$ is also such a vector. To make this more clear, let $\alpha a = b \in R$ and substitute it:

$\alpha \vec{x} = (\alpha a, 2 (\alpha a), 3 (\alpha a))=(b, 2 b, 3b)$. Clearly, $\alpha \vec{x}$ is also in S.

Because $\alpha$ is any arbitrary real rumber, and $\vec{x}$ is any arbitrary vector in S, we have shown that S is closed under scalar multiplication.

Hope that makes sense.
• April 6th 2009, 03:16 AM
U-God
Ok, thanks so far.
For the second one $(x,y,z) \mid 2x - 3y + 4z = 1$

I can see that if $x = \frac{1}{2}$, $y = 0$ and $z = 0$ then the the equation is valid, therefore it is not an empty set.

if I took $\vec{a} = (x_1,y_1,z_1)$ and $\vec{b} = (x_2,y_2,z_2)$

Then $\vec{a} + \vec{b} = (x_1 + x_2, y_1 + y_2, z_1 + z_2)$

This still has to satisfy the condition: $2x - 3y + 4z = 1$

so I can say that $2(x_1 + x_2) - 3(y_1 + y_2) + 4(z_1 + z_2) = 1$

So is this subset of $R^3$ only a subspace under certain values of $x_1, x_2, y_1, y_2, z_1, z_2$?
If so how do I define this?
Thanks

Edit:
If I consider $\vec{a}$ and $\vec{b}$ to be on the plane $2x - 3y + 4z = 1$, then any sum of $\vec{a}$ and $\vec{b}$ will also lie on the plane. Therefore closed under vector addition. I know this isn't sufficient, but I think I may be on the right track..

Edit #2: back to $\vec{a} = (x_1,y_1,z_1)$ and $\vec{b} = (x_2,y_2,z_2)$
From this, $\vec{a} = 1$ and $\vec{b} = 1$ therefore, $\vec{a} + \vec{b} = 2$ which is not allowed as $2x - 3y + 4z = 1$
Which means that $2(x_1 + x_2) - 3(y_1 + y_2) + 4(z_1 + z_2) != 1$ so therefore it is not closed under vector addition!

Please tell me that this one is right.
Thanks,
• April 7th 2009, 05:32 PM
scorpion007
Let $S=\lbrace (x,y,z) \mid 2x - 3y + 4z = 1 \rbrace \subset R^3$.

The quick way to prove that it is not a subspace is by showing that $\vec{0} \notin S$. I.e. $2(0) - 3(0) + 4(0) \neq 1$.

But just to show that it isn't closed under vector addition:

Let $\vec{a} = (x_1,y_1,z_1), ~\vec{b} = (x_2,y_2,z_2)$ be any vectors in S. Therefore we know that $2x_1-3y_1+4z_1 = 1$ and likewise $2x_2-3y_2+4z_2 = 1$.

Then, $\vec{a}+\vec{b} = 2(x_1 + x_2) - 3(y_1 + y_2) + 4(z_1 + z_2)$.
Expanding and grouping terms, we get:
$\vec{a}+\vec{b} = \underbrace{(2x_1 -3y_1 +4z_1)}_{1} + \underbrace{(2x_2 - 3y_2 + 4z_2)}_{1} = 2$

$\therefore \vec{a}+\vec{b} \notin S$, and as such S is not closed under addition.