Question

If we have n linearly independent vectors for dimension n vector space, those n vectors span the space. How can I prove it?

Answer 1

To show the n vectors span the space, it is necessary to show that ANY vector in that space can be expressed as a linear combination of those n vectors. In other words, if w is ANY vector in the space, the equation

w= C1V1 + C2 V2 + C3V3 + ...+CnVn must have solutions for the C1, C2...Cn. Writing this vector equation in terms of the components of the vectors will give a system of linear equations:

w1 = C1V11 + C2V21 + ....+ CnVn1
w2 = C1V21 + C2V22 + ... + CnVn2
. . . .
wn = CnVn1 + CnVn2 + ... + CnVnn

This system of n Linear Equations will have solutions (i.e. consistent) in C1, C2,...Cn only and only if the coefficient matrix, (below) is INVERTIBLE.

V11 V21 Vn1
V21 V22 Vn2
. . . .
Vn1 Vn2 Vnn.

To prove that this matrix is indeed INVERTIBLE, we make use of the given fact that the n vectors are Linearly Independent (L.I.). L.I. means that the only scalars that satisfy the equation
0 = k1V1 + k2V2 + ... +knVn -----(1)

are k1=k2=k3....=kn=0

Writing (1) out in terms of the components of the vectors gives the system of linear equations:below:

0 = k1V11 + k2V21 + ....+ knVn1
0 = k1V21 + k2V22 + ... + knVn2

0 = knVn1 + knVn2 + ... + knVnn

L.I. requires this Homogeneous System to have only Trivial Solution (k1=k2=k3=...kn=0), which means the coefficient matrix is INVERTIBLE.
Hence, the proof. Q.E.D.

Answer 2

hotjava, Thanks for the such a neatly written proof.

However I don't get why we can assume a vector in V(your w) is n-component.
as expressed in wn = CnVn1 + CnVn2 + ... + CnVnn

I looked up another book and found a proof.

It doesn't assume n-component for w.
(I guess it's possible to prove somehow that vectors of dimension n space is composed of n component?)

For someone interested, i'll leave the proof sketch.
1. if {x1,x2..xm} spans V, then every set of vectors with more than m vectors cannot be linearly independent.(required to prove 2)
2. Any linearly independent set in V can be extended to a basis by adding more vectors if necessary. (1)
Any set of vectors that spans V can be reduced to a basis by discarding vectors if necessary. (2)
3. Let V be a vector space of dimension n, Then any set of n linearly independent vectors is a basis for V. (proof uses 2-(1))

Answer 3

The question says we are dealing with a n-dimensional vector space.. Any vector in a n-dimension space has n components.

Answer 4

hotjava, that seems plausible, but I learned 'dimension' is number of basis of a space. How do I go from the definition of dimension to the "n-dimension space has n components"?

Answer 5

A n-dimensional space refers to Rn, which can be described as a n-tuple i.e. a vector with n components.