# PROOF: Independent vectors and spanning vectors

#### Ella087

Proof:
1. why you need at least m vectors to span a space of dimension m.
2. If m vectors span an m-dimensional space, then they form a basis of the space.

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#### ice109

do you know how to do proofs?

#### HallsofIvy

Homework Helper
What does it mean to say that a vector space has finite dimension? (NOT just that "its dimension is finite. You have to have 'finite' dimension before you can define dimension.) What is the definition of "dimension n".

#### JasonRox

Homework Helper
Gold Member

And reading some proofs of some easy theorems. You'll see how to apply definitions to prove statements like you have.

#### mathwonk

Homework Helper
these are proved in my free linear algebra notes on my web page. the full details are not given, but i believe the proofs can be filled in without huge difficulty, if you try and understand the definitions.

#### transgalactic

lets start from the definition that a basis consists only from
independant vectors
now in order to span some space in "n" dimention you need to have "n" independant vectors
which are a basis for this dimention.

#### tomyus

Find the independance

Find if these vectors are lineary independant :..

U=(1 2 9) . v=(2 3 5) .

I know the condition of the lineary independant is

au+bv=0 but how I can use this condition here .,.,.,

Please If you know the answer u can send it at >> ahmedtomyus@yahoo.com

Last edited:

#### transgalactic

noooooooooooooo
dont use that formula
stack them one upon the other as matrix and make a row reduction
if you dond have a line of zeros in the resolt then they are independant

#### HallsofIvy

Homework Helper
Find if these vectors are lineary independant :..

U=(1 2 9) . v=(2 3 5) .

I know the condition of the lineary independant is

au+bv=0 but how I can use this condition here .,.,.,

Please If you know the answer u can send it at >> ahmedtomyus@yahoo.com
Normally I would not respond to a question that has nothing to do with the original question but since transgalactic responded to it...

noooooooooooooo
dont use that formula
stack them one upon the other as matrix and make a row reduction
if you dond have a line of zeros in the resolt then they are independant
Why not? au+ bv= a(1, 2, 9)+ b(2, 3, 5)= (a+2b, 2a+ 3b, 9a+ 5b)= (0, 0, 0) seems easy enough. We must a+ 2b= 0, so a= -2b. Then 2a+ 3b= -6a+ 3b= -3b= 0. b must equal 0, so a= -2b= 0 also. Since a and b must both be 0 the two vectors are, by definition, independent.

Yes, if we have 10 vectors or 10000 then "row reduction" would be simpler but I believe it is better practice to use the basic definitions.

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