Basis of Vector Space

• December 6th 2013, 02:43 PM
davidciprut
Basis of Vector Space
Hi, I have some questions about basis.

1-)Let's say w1,w2,......,wL is in the Span(U), (U={u1,u2,.....,uK} , u1,u2,.....,uK are the basis of U)
If L>K then w1,w2,......,wL is linearly independent.

I understand the statement intuitively very good, but can someone give me a formal proof? I don't feel fully convinced.

2-) If we have a Vector Space U, and its basis consists of , lets say n vectors, all other bases that I will find for the vector space U will consist of n vectors? I can't find more linearly independent vectors that spreads the whole vector space that is also the base?

Thanks you
• December 6th 2013, 03:51 PM
HallsofIvy
Re: Basis of Vector Space
Quote:

Originally Posted by davidciprut
Hi, I have some questions about basis.

1-)Let's say w1,w2,......,wL is in the Span(U), (U={u1,u2,.....,uK} , u1,u2,.....,uK are the basis of U)
If L>K then w1,w2,......,wL is linearly independent.

I understand the statement intuitively very good, but can someone give me a formal proof? I don't feel fully convinced.

Perhaps you do not understand it! What you wrote is NOT true- unless you simply mistyped it- If L> K then w1,w2,......,wL is linearly dependent, not independent.

To see that use the definition of "dependent-independent". The vectors w1,w2,......,wL are linearly dependent if and only if there exist numbers, a1, a2, ..., aL, NOT all 0, such that a1w1+ a2w2+ ...+ a[sub]L[sub]wL= 0.

Since u1, u2, ..., uK form a basis for U, they span the space: we can write each of the "w" vectors as a linear combination of the "u" vectors:
w1= b21u1+ b22u2+ ...+ b2LuL
w2= b21u1+ b22u2+ ...+ b2LuL
...
wK= bK1u1+ bK2u2+ ...+ bKLu[sub][L/sub]

And we can replace all those "w"s in the original linear combination to get
a1w1+ a2w2+ ...+ a[sub]L[sub]wL= a1(b21u1+ b22u2+ ...+ b2LuL)+ a2(b21u1+ b22u2+ ...+ b2LuL)+ ... + aL(bK1u1+ bK2u2+ ...+ bKLu[sub][L/sub])
= (a1b11+ a2b21+ ...+ aLbL1)u1+ (a1b12+ a2b22+ ...+ aLbL2)u2+ ... + (a1b1K+ a2b2K+ ...+ aLbLK)uK)uK= 0.

Since the u vectors are a basis, they are independent so each of the coefficients must be 0:
a1b11+ a2b21+ ...+ aLbL1= 0
a1b12+ a2b22+ ...+ aLbL2= 0
...
a1b1K+ a2b2K+ ...+ aLbLK= 0

Now, since we have "K" equations in "L" unknowns, and L> K, we could, for example, solve the equations for specific values of K unknowns, in terms of the other L- K unknowns. Any particular, we could choose one or more of the L- K unknowns to be non-zero proving that the vectors are dependent.

Quote:

2-) If we have a Vector Space U, and its basis consists of , lets say n vectors, all other bases that I will find for the vector space U will consist of n vectors? I can't find more linearly independent vectors that spreads the whole vector space that is also the base?

Thanks you
If there exist a basis for a vector space, consisting of n vectors, then, by definition of basis, they span the space and are independent.

To show that a set with m> n vectors cannot be independent, write a linear combination equal to 0. Replace each of the m vectors with its expression as a linear combination of the n basis vectors and combine coefficients of the same basis vector. That gives a linear combination of the basis vectors equal to 0 so each coefficient of each basis vector is equal to 0. As above, that gives n equations in more than n unknowns which must have a non-trivial solution.

To show that a set with m< n vectors cannot span the space, assume that one does, then do effectively the same thing but writing a linear combination of the basis vectors in terms of the m vectors that span the space. That shows that the original set of n vectors cannot be independent, contradicting the definition of a basis.
• December 6th 2013, 04:30 PM
davidciprut
Re: Basis of Vector Space
Quote:

Originally Posted by HallsofIvy
Perhaps you do not understand it! What you wrote is NOT true- unless you simply mistyped it- If L> K then w1,w2,......,wL is linearly dependent, not independent. .

Ups, my fault, I didn't notice , I meant dependent!
• December 6th 2013, 06:11 PM
Plato
Re: Basis of Vector Space
Quote:

Originally Posted by davidciprut
Ups, my fault, I didn't notice , I meant dependent!

If $U$ is spanned by $K$ independent vectors, then we we say $\text{dim}(U)=K$.

Therefore, if $L>K$ then any set of $L$ vectors must be dependent.