# A question to vector subspace

1. Jul 20, 2007

### uiulic

Anton, H. Elementary linear algebra (5e, page 156) says:

If W is subset of a vector space V, then W is a subspace of V if and only if the following TWO conditions hold

1) If u and v are vectors in W, then u+v is in W
2) If k is any scalar and u is any vector in W, then ku is in W

However, Lang, S. Introduction to linear algebra (UTM, page 91) adds another condition besides the above two mentioned conditions, which is
3) The element O of V is also an element of W.

wiki has the same treatment as Lang' book. It seems that 3) can be obtained from 2) by setting k=0. The question is whether 3) is necessary? Is there anything I miss in understanding the books and wiki?

2. Jul 20, 2007

### cristo

Staff Emeritus
What do you mean by "the element O"? Do you mean the zero vector? If so, then your condition 3) seems unnecessary, since one can use 2) to show that the zero vector is in W, as you say. I don't see where wiki says states this third condition: I see that it defines a subspace to be a subset W of V that is closed under vector addition and scalar multiplication.

3. Jul 20, 2007

### uiulic

Yes. My "O" means zero vector.

Wiki's theorm at
http://en.wikipedia.org/wiki/Linear_subspace

Theorem: Let V be a vector space over the field K, and let W be a subset of V. Then W is a subspace if and only if it satisfies the following 3 conditions:

1)If the zero vector, θ, is in W.
2)If u and v are elements of W, then the sum u + v is an element of W;
3)If u is an element of W and c is a scalar from K, then the scalar product cu is an element of W;

wiki says property 1) ensures W is nonempty, which is what I don't understand. Also Serge Lang's book not only clearly adds the zero vector condition in the statement of the subspace theory, but also in his examples.

4. Jul 20, 2007

### cristo

Staff Emeritus
Ahh, ok, well the first definition should probably have the condition W a nonempty subset of V (see the subspace section here)

Clearly, a subspace of a vector space is a vector space on its own, and the empty set would not satisfy the vector space axioms.

5. Jul 20, 2007

### uiulic

To cristo, you mentioned

"I don't see where wiki says states this third condition: I see that it defines a subspace to be a subset W of V that is closed under vector addition and scalar multiplication."

Where dis you see this in wiki? Shouldn't subspace be a vector space, which indicates "closed" itself?

Thanks

6. Jul 20, 2007

### uiulic

But the other conditions clearly mention that there ARE elements in the subset?

7. Jul 20, 2007

### cristo

Staff Emeritus
Yes, but checking the subspace "axioms" for a nonempty subset of a vector space is normally the quick way of checking whether W is a vector space, rather than checking all the vector space axioms.
Well, technically, they don't do they? The conditions say if .... not, there exist...

8. Jul 20, 2007

### HallsofIvy

Staff Emeritus
A "theorem" is NOT a definition! When defining things we try to keep it as simple as possible. The definition of "subspace" is just what Anton gives.

It then follows that the 0 vector must be in a subspace (as well as "if v is in the subspace then so is -v". Those are results of the definition, not part of the definition.

9. Jul 20, 2007

### uiulic

Yes. Your explanation may have explained why some authors choose the "zero vector" condition for the subspace. But wiki uses "if and only if",and it seems confusing for adding one obvious "dependent" requirement.

10. Jul 20, 2007

### uiulic

It should be pointed out that Anton also states it is the theorem in fact.

11. Jul 20, 2007

### uiulic

I always suspect the conditions for the definition of VECTOR SPACE (8 or 10 axioms). Are they independent of each other? As a definition, a dependent requirement will not cause trouble in mathematics, I think.

Halmos stated the requirements for a VECTOR SPACE are not claimed to be logically independent? (There had been a long history for Eucli's fifth axiom...) Could you give one example for confirming Halmos' words?

12. Jul 21, 2007

### fopc

The standard set of VS axioms is not an independent set.
The commutative axiom for vector addition can be proved from the remaining axioms.

I'd say from a practical point of view, it's of little significance.
I guess that's why Halmos made nothing more than a passing remark about it.
Birkhoff and MacLane also comment on it.

You can see the same thing in the typical axiom set for Boolean algebra (the algebraic structure).
There, it's the axioms of associativity.

Last edited: Jul 21, 2007
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