Question

The dimension of a vector space V is n.
Prove: If \(S=\left \{\mathbf{v_1},\mathbf{v_2},...,\mathbf{v_n}\right \}\) is linearly independent, then S is a basis for V.

Answer 1

what is your definition of a basis?

Answer 2

A set B is a basis for V if two conditions are satisfied:
1. B is linearly independent.
2. The span of B is V.

Answer 3

well, condition 1 is given as part of the information

Answer 4

How do you show that a set of "n" linearly independant vectors spans a dimension of "n"?

Answer 5

Well, I should prove that for any \(\mathbf{v}\in V\), it can be written as a linear combination of the vectors in S.

Answer 6

that sounds like a good approach

Answer 7

we know the zero vector is in it since they are stated to be LI to start with ...

Answer 8

since S is linearly independant; it is row equaivalent to the nxn identity matrix; which spans R^n

Answer 9

But that only applies to R^n. It doesn't cover arbitrary vector spaces.

Answer 10

how do you define "dimension of a vector space" ?

Answer 11

brb, ill have to run upstairs and get my david lay book .....

Answer 12

The dimension of a vector space is the number of vectors in any basis. We have learned that if a basis for V has n vectors, then every other basis for V has n vectors.

Answer 13

the dimension of a vector space V is the number of linearly independant vectors that form a basis for V

Answer 14

since S is linearly independant with "n" vectors, it is row equavilent to the nxn identity matrix which is a basis for V. Im pretty sure thats aribtrary ....

Answer 15

since the dimension of vector V is given to be "n"; and the nxn identity matrix forms a basis for an n-dimension Vector space .....

Answer 16

S being row-equivalent to nxn identity matrix which forms a basis for n-dimensional vector space means that S also forms a basis for the same vector space?

Answer 17

Basis Theorem: Let \(V\) be an \(n\)-dimensional vector space , \(n\)>=1. Any linearly independant set of exactly \(n\) elements in \(V\) is automatically a basis for \(V\). Any set of exactly \(n\) elements that spans \(V\) is automatically a basis for \(V\).

Proof: By thrm 11 (its the one before this one in the book); a LI set \(S\) of \(n\) elements can be extended to a basis for \(V\). But that basis must contain exactly \(n\) elements, since dim \(V\)=\(n\). So \(S\) must already be a basis for V. Now spose that \(s\) has \(n\) elements and spans \(V\). Since \(V\) is nonzero, the Spanning Set Thrm implies that a subset of \(S'\) of \(S\) is a basis of \(V\). Since dim\(V\)=\(n\), \(S'\) must contain \(n\) vectors. Hence, \(S=S'\).

from the textbook

Answer 18

"A LI set \(S\) of \(n\) elements can be extended to a basis for \(V\)."
So this statement says that if we wanted to add vectors to S so that it forms a basis for V, we could, but we don't really need to because V has a dimension of n?

Answer 19

my prior comments: 2 matrixes are row equivalent if you can use elementary row operations to convert the elements of one into the elements of the other.

Since a linearly independant set of n vectors row reduces to the nxn identity matrix; then they are row equivalent. The I nxn matrix is automatically a basis for n-dimensional vector space

Answer 20

thrm 11, the one before this in the book :), Any LI set in V can be expanded, if necessary, to form a basis for V. theres another paragraph of proof below it :)

Answer 21

lets assume a vector space H is dim 5

H contains [1 0 0 0 0] = v1
H contains [0 1 0 0 0] = v2
H contains [0 0 1 0 0] = v3
H contains [0 0 0 1 0] = v4
H contains [0 0 0 0 1] = v5

we cannot introduce any other vector and this remain LI can we?

Answer 22

And is that because we said that H has dimension of 5?

Answer 23

yes

Answer 24

And this argument can certainly be applied to any vector space V.
Thanks for the help!

Answer 25

good luck :)

Answer 26

i saw later that my Identity nxn was not general enough :)

M{rxc} can be linearly independant so long as r >= c

in my last example, lets assume dimH = 3, but still use the 5-tuples such that:

H contains [1 0 0 0 0] = v1
H contains [0 1 0 0 0] = v2
H contains [0 0 1 0 0] = v3

v1,v2,c3 are still LI, giving us a 5x3 basis that spans all of H.