Re: Basis of Vector Space

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Originally Posted by

**davidciprut** Hi, I have some questions about basis.

1-)Let's say w_{1},w_{2},......,w_{L} is in the Span(U), (U={u_{1},u_{2},.....,u_{K}} , u_{1},u_{2},.....,u_{K }are the basis of U)

If L>K then w_{1},w_{2},......,w_{L }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 w_{1},w_{2},......,w_{L } is linearly **dependent**, not independent.

To see that use the definition of "dependent-independent". The vectors w_{1},w_{2},......,w_{L } are linearly **dependent** if and only if there exist numbers, a_{1}, a_{2}, ..., a_{L}, NOT all 0, such that a_{1}w_{1}+ a_{2}w_{2}+ ...+ a[sub]L[sub]w_{L}= 0.

Since u_{1}, u_{2}, ..., u_{K} 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:

w_{1}= b_{21}u_{1}+ b_{22}u_{2}+ ...+ b_{2L}u_{L}

w_{2}= b_{21}u_{1}+ b_{22}u_{2}+ ...+ b_{2L}u_{L}

...

w_{K}= b_{K1}u_{1}+ b_{K2}u_{2}+ ...+ b_{KL}u[sub][L/sub]

And we can replace all those "w"s in the original linear combination to get

a_{1}w_{1}+ a_{2}w_{2}+ ...+ a[sub]L[sub]w_{L}= a_{1}(b_{21}u_{1}+ b_{22}u_{2}+ ...+ b_{2L}u_{L})+ a_{2}(b_{21}u_{1}+ b_{22}u_{2}+ ...+ b_{2L}u_{L})+ ... + a_{L}(b_{K1}u_{1}+ b_{K2}u_{2}+ ...+ b_{KL}u[sub][L/sub])

= (a_{1}b_{11}+ a_{2}b_{21}+ ...+ a_{L}b_{L1})u_{1}+ (a_{1}b_{12}+ a_{2}b_{22}+ ...+ a_{L}b_{L2})u_{2}+ ... + (a_{1}b_{1K}+ a_{2}b_{2K}+ ...+ a_{L}b_{LK})u_{K})u_{K}= 0.

Since the u vectors are a basis, they are independent so each of the coefficients must be 0:

a_{1}b_{11}+ a_{2}b_{21}+ ...+ a_{L}b_{L1}= 0

a_{1}b_{12}+ a_{2}b_{22}+ ...+ a_{L}b_{L2}= 0

...

a_{1}b_{1K}+ a_{2}b_{2K}+ ...+ a_{L}b_{LK}= 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**.

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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.

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 w_{1},w_{2},......,w_{L } is linearly **dependent**, not independent. .

Ups, my fault, I didn't notice , I meant dependent!

Re: Basis of Vector Space

Quote:

Originally Posted by

**davidciprut** Ups, my fault, I didn't notice , I meant dependent!

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

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