# Three properties of subspaces

1. Jan 6, 2005

### matrix_204

I had a question regarding subspaces.
Given vectors $$(a,b,c,d)$$ s.t. $$\left{\mid}\begin{array}{cc}a&b\\c&d\end{array}\right{\mid}=0$$
a supspace of $$\Re^4$$?

Though i kno the answer is yes, but i dont understand like it looks to me that it uses one of the three properties of subspaces; the zero vector one i m guessing. But could someone plz explain this to me?

2. Jan 6, 2005

### Hurkyl

Staff Emeritus
Don't panic! Stop, take a breath, and collect your thoughts.

You seem to be asking for help in understanding why this particular class of vectors form a subspace. So, you should look at each of the conditions that a subset be a subspace one at a time and see if they apply.

P.S. I think you were trying to write

$$\left| \begin{array}{cc} a&b \\ c&d \end{array} \right|$$

3. Jan 6, 2005

### matrix_204

lets say v1=(1,0,0,0) and v2=(1,1,1,1), by adding the two vectors, u get (2,1,1,1) and it turns out 2(1)-1(1)doesn't equal 0, so condition one fails, right?

4. Jan 6, 2005

### Hurkyl

Staff Emeritus
If the set is not closed under addition*, then it is not a subspace.

* don't refer to this simply by number -- the only people who will understand you are those who know your textbook and will look it up

5. Jan 6, 2005

### matrix_204

so it seems like it is not closed under addition, but the teacher said as a hint that it is a subspace of R^4.

6. Jan 6, 2005

### Hurkyl

Staff Emeritus
Well, if you're confident in your analysis, then one of two things must be true:

(b) You have misunderstood the problem.

I'll save you a little bit of headache -- if I understand your initial post, then you have correctly analyzed the problem you stated.

7. Jan 6, 2005

### matrix_204

ok i m pretty sure i m wrong, but i will restate the problem in the way the teacher said.
Is the set of all vectors (a,b,c,d) such that $$\left|\begin{array}{cc}a&b \\c&d\end{array}\right|=0$$ a subspace of R^4?

8. Jan 6, 2005

### Hurkyl

Staff Emeritus
Have you produced two vectors in that set whose sum is not in that set?
Is that sufficient to prove the set is not a vector space?

9. Jan 6, 2005

### matrix_204

well now i understand a bit more, and checked that there is no vector whose sum is not in the set. So it satisfies the condition of addition. Similarly satisfies the second condition of multiplication, but now I m confused in how to check it using the Zero vector condition. Can you give me an example of checking something using the zero vectors?

10. Jan 6, 2005

### Hurkyl

Staff Emeritus
You sure? What was wrong with your example of:

<1, 0, 0, 0> + <1, 1, 1, 1> = <2, 1, 1, 1>

Anyways, I'm not sure what you mean by "zero vector condition" -- as I remember, the only thing you need to check is closure under addition and multiplication.

Note that closure under multiplication proves that it contains the zero vector (multiply by 0).