# Do planes stop, or are they ever expanding? [closed]

I am trying to understand sub-spaces in linear algebra and one of the rules mentions if $$W$$ is my subspace then if $$k$$ is any scalar and $$\bf u$$ is any vector in $$W$$ then $$k\bf u$$ is in $$W$$.

I am unsure how this works ? Wouldn't there be a number I could use for the scalar that would make $$u$$ so big it would leave the subspace ? i.e if my subspace was a plane in $$\Bbb R^3$$ wouldn't the vector off the edge of my plane? or are plane's edges ever expanding ? as in do the edges of a plane keep going in the direction they are going or do they stop ?

hope this makes sesne

• Think about it. Does it really have "edges"? Commented May 20, 2015 at 4:23
• Is there an upper bound on $\mathbf{k}$? Commented May 20, 2015 at 4:24
• maybe sides is a better term ? Commented May 20, 2015 at 4:24
• There aren't any sides in a plane. It goes as far as you want. Commented May 20, 2015 at 4:24
• graydad even if there's an upper bound on k, if there's a k > 1 then you can say x is in S, so kx is too, then k(kx)... to infinite and beyond! Commented May 20, 2015 at 4:25

A plane in $\Bbb{R}^3$ has no edges, even though it seems like they do when we draw them. Your idea of "make it so big it would leave the subspace" is in direct opposition with the rule you just wrote.

• ohh right, yeah the rule had me confused when looking at the pictures in the textbooks Commented May 20, 2015 at 4:26

A plane in $R^3$ is infinitely large from the begining, it doesn't "expand" at all. There is no way then that any scalar multiplication can make it go outside of it

According to the definition of a subspace, what you have described is impossible. Namely, a subspace has to be close under scalar multiplication. In other words, if you take a vector $\bar v$ from a subset of a vector space, and you see that $k\bar v$ is not in that subset anymore, then by definition, that subset cannot possibly be a subspace.

• I am unsure how this could be now. If a plane is infinitely large at the out set and v is already in it , then I am unsure how kv could leave it ? Commented May 20, 2015 at 4:31
• @user3754366 No, it couldn't. But that doesn't mean the plane is a subspace. Commented May 20, 2015 at 4:35
• hmm my mind is boggling...I'd thought the rule was if u and v are vectors in W, then u + vis in W and if k is any scalar and u is any vector in W, then ku is in W. Initially reading these makes it seem like it'd never leave the plane ..any clues as to what concept I am not grasping ? Commented May 20, 2015 at 4:38
• @user3754366 I'm not sure what confuses you. But keep in mind that the plane has to pass through the origin to be a subspace. If you just take some random plane in $\mathbb{R}^3$, it may not contain the zero vector. Commented May 20, 2015 at 4:40
• ohh so even if kv didn't leave the plane, if the plane never went through the origin it wouldn't be a subspace anyway Commented May 20, 2015 at 4:41

To start off, you asked a fairly good question - namely because I haven't though about subspaces in the way you're asking.

In any case, I'll take a shot at your question:

To make things simple, I'll assume the vector space we're working with to be in $R^3$ like you mentioned. Furthermore, I'll choose a plane - specifically the $X-Y$ plane - which is $\in R^2$.

Since pictures make potentially abstract ideas easier to comprehend, I've included the following:

(I pulled it off Google somewhere).

As you can see, we have our vector space $R^3$, and our subspace - which we defined to be the $X-Y$ plane. Now things can get a little more abstract. A vector space, as you're obviously aware being that you asked the question, is defined to be a set that satisfies the following axioms (courtesy of Wolfram):

1. Commutativity:

X+Y=Y+X.

2. Associativity of vector addition:

(X+Y)+Z=X+(Y+Z).

3. Additive identity: For all X,

0+X=X+0=X.

4. Existence of additive inverse: For any X, there exists a -X such that

X+(-X)=0.

5. Associativity of scalar multiplication:

r(sX)=(rs)X.

6. Distributivity of scalar sums:

(r+s)X=rX+sX.

7. Distributivity of vector sums:

r(X+Y)=rX+rY.

8. Scalar multiplication identity:

1X=X.

Now that you have a vector space defined, you can create vector spaces within vector spaces - also known as a subspace. Being that you already know that you have a valid vector space, the only two conditions that must be satisfied are closure under vector addition, and closure under scalar multiplication. Formally stated:

1. W is nonempty, W≠∅.
2. If x∈W and y∈W, then x+y∈W.
3. If α∈C and x∈W, then αx∈W.

(Note that some people also formally include the condition that your subspace is non-empty).

After all this, it is worth noting again that subspaces and vector spaces are very much definition driven. In other words, a set is only a subspace if and only if it satisfies the above axioms.

Without looking at the image, we can see that if we have a subspace, then we know that any scalar we pick must be part of that subspace (i.e. it will never be "too" big as your wondering). Why? Because we have a subspace, which means that any vector that's an element of my subspace multiplied by a scalar $k$ will still be part of my subspace).

Considering the image, a basis for our vector space, $R3$, can simply be the standard basis: $\{(1,0,0)^T, (0,1,0)^T, (0,0,1)^T\}$. A basis for our subspace, then, would be $\{(1,0,0)^T, (0,1,0)^T\}$. As you can probably intuitively know, your X-axis can go on forever, your Y-axis can go on forever, and so forth. So your planes never end in this case (unless you define them to).

Anyway, what's worth keeping in mind are the definitions of a vector space and subspace (especially once you start talking about function spaces). If you have a subspace, then you already know $ku$ will still be in the subspace.

Hope that helped!

You have not selected an answer yet, so here's a short explanation of why this is impossible strictly from a logic standpoint. We have the claim

For all $k\in \mathbb{N}$ and $\mathbf{u} \in W$, we have $k\mathbf{u} \in W$

And you are also trying to introduce the claim

There exists $k\in \mathbb{N}$ along with a $\mathbf{u}\in W$ such that $k\mathbf{u} \notin W$

These two claims contradict each other, in the same way the claim "$x$ is even and not even" is a contradiction. Since the first claim is part of the definition of a subspace, we go with it, and drop the second claim. Hence the second claim is necessarily false.