# Infinitely many subspaces in R3 ?

1. Mar 16, 2006

### Mimi

infinitely many "subspaces" in R3 ?

In R3, there are zero, 1, 2, 3 dimensional subspaces. But how can I express them with 'specific' example, using variables x,y,and z?

2. Mar 16, 2006

### JasonRox

There are infinite many lines through the origin.

That's a hint.

3. Mar 17, 2006

### HallsofIvy

Staff Emeritus
And there are infinitely many planes containing the origin! But R3 has only one 3 dimensional subspace.

If by "specific example" you mean a specific example of each: what is the equation of a line through the origin? What is the equation of a plane containing the origin? And as I said, you don't have many choices for an example of a 3 dimensional supspace.

4. Mar 17, 2006

### benorin

In addition, know that every subspace of R^n contains the origin.

5. Mar 17, 2006

### JasonRox

That's false.

You are assuming the zero vector to be the origin. It doesn't have to be.

6. Mar 18, 2006

### benorin

I did mean to state this only for the usual definitions of scalar multiplication and vector addition; in such a case, R^3 has precisely these subspaces: the trivial ones, namely, the set containing the zero vector and R^3 itself, (which are the zero and 3 dimensional subspaces); each line containing the origin is a subspace of dimension 1 (and there are no others of dimension 1); and lastly, each plane containing the origin is a subspace of dimension 2 (and there are no others of dimension 2).

7. Mar 18, 2006

### JasonRox

That's better.

I got a True/False question wrong once because I answered it too quickly and it was regarding this. Not directly, but in general.

So, for the sake of being concise, I always try to add the conditions when I make a comment or statement. I forget sometimes, but that's normal.

The reason why I pointed it out here is because the thread starter is relatively new to linear algebra, and he/she should get a feel of what a vector space is. People get mistaken that a vector space actually has 0 as the zero vector all the time.

8. Mar 18, 2006

### HallsofIvy

Staff Emeritus
The original post said "subspace of Rn". That is sufficient to conclude that the 0 vector is (0,0). I'm not sure what you mean by "People get mistaken that a vector space has 0 as the zero vector all the time". Are you thinking about a case in which there is an unusual definition of vector addition? Precisely because "the thread starter is relatively new to linear algebra" I think bringing up situations like that when it was clearlys stated that this is in Rn will be confusing.