## PROOF: Independent vectors and spanning vectors

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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 do you know how to do proofs?
 Recognitions: Gold Member Science Advisor Staff Emeritus 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".

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## PROOF: Independent vectors and spanning vectors

Try reading some definitions in your textbook.

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

 Recognitions: Homework Help Science Advisor 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.
 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.
 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
 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

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 Quote by tomyus 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...

 Quote by 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
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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