# Thread: Tricky first year proof

1. ## Tricky first year proof

let V be a vector space and suppose B is a basis for V. show that any vector in V be written uniquely as a linear combination of elements from B.

Starters/pointers/hints would be great

Nic

2. Basis B is the largest set of linearly independent vectors in V. Number of elements of that set represents the dimension of V. Take any vector v in V. Then v together with the elements of B forms a set of linearly dependent vectors. Form a linear combination and it can be zero with not all coefficients equal to zero.

See where that gets you.

3. Suppose otherwise, and you’ll end up enlarging your basis.

4. Don't forget to prove uniqueness. Suppose there are two ways to express a single vector v and look for contradiction.

5. Originally Posted by MathoMan
Basis B is the largest set of linearly independent vectors in V. Number of elements of that set represents the dimension of V. Take any vector v in V. Then v together with the elements of B forms a set of linearly dependent vectors. Form a linear combination and it can be zero with not all coefficients equal to zero.

See where that gets you.
Ok, so I understand the description, but in terms of actually defining the vectors. For instance, you say 'take any vector v in V, would you define it is [A1+A2....An]? and then how would you define the elements of B?

I understand what I have to do, just can't get it down on paper so to say.

6. First, what is the definition of "basis" you are using? In some texts, it is exactly what you are trying to prove. Probably your definition is what Mathoman said- a largest independent set of vectors. Although, be careful about saying the basis is the largest set- there exist an infinite number of bases, all having the same size.

Since this is "first year", can we assume that the vector space is finite- dimensional? In that case you can write your basis, B, as, say, $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$.

Let v be any vector in the vector space. Since B is a largest set of independent vectors, $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$ is NOT independent. Use the definiton of "dependent" to show that v can be written as a linear combination of vectors in B.

Now uniqueness. Suppose there were some vector, v, which could be written as two different linear combinations of vectors in B. That is, $v= a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n$ and $v= b_1v_1+ b_2v_2+ \cdot\cdot\cdot+ b_nv_n$ where, for at least some index i, $a_i\ne b_i$. Subtract the two equations to get a contradiction to the fact that the set is independent.

7. Originally Posted by Tikoloshe
Suppose otherwise, and you’ll end up enlarging your basis.
Tikoloshe stated the most elegant way of doing this.

B is basis for V. Then B is largest possible set of vectors all being linearly independent of each other. If arbitrarily chosen vector v in V cannot be expressed as a linear combination of elements in B, than it follows it is linearly independent of all elements in B. Hence v together with elements in B forms another set B' of linearly independent vectors that is larger then B (obviously B is contained in B'). That contradicts with the fact that B IS basis for V and as such B is largest posible set of linearly independent. Hence one must reject the assumption that v cannot be expressed as a linear combination of elements in B, thus leaving the oposite as true: any v in V can be expressed as a linear combination of elements from B.

8. Hey. By the definition of basis, any vector in the space can be expressed as a linear combination of the elements of V. Suppose that this representation was not unique:

ie $c_1v_1+...+c_nv_n=d_1v_1+...+d_nv_n$

Then we have $0=(c_1-d_1)v_1+...+(c_n-d_n)v_n$. Then, kick in linear independence.