Thread Closed

PROOF: Independent vectors and spanning vectors

 
Share Thread Thread Tools
Mar13-08, 07:42 PM   #1
 

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.
 
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> 'Whodunnit' of Irish potato famine solved
>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change
>> Curiosity Mars rover drills second rock target
Mar13-08, 08:30 PM   #2
 
do you know how to do proofs?
 
Mar14-08, 06:41 AM   #3
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff 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".
 
Mar14-08, 12:00 PM   #4
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help

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.
 
Mar16-08, 01:18 PM   #5
 
Recognitions:
Homework Helper Homework Help
Science Advisor 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.
 
Mar18-08, 03:10 PM   #6
 
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.
 
Mar19-08, 06:11 AM   #7
 
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
 
Mar19-08, 03:50 PM   #8
 
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
 
Mar20-08, 10:42 AM   #9
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff Staff Emeritus
Quote by tomyus View Post
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 View Post
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.
 
Thread Closed
Thread Tools


Similar Threads for: PROOF: Independent vectors and spanning vectors
Thread Forum Replies
Expressing cartesian unit vectors in terms of spherical unit vectors General Math 9
About polar vectors and pseudo vectors Classical Physics 10
The dimension of the span of three linearly independent R^3 vectors Calculus & Beyond Homework 6
Position Vectors, Velocity Vectors, and Acceleration Vectors Introductory Physics Homework 3
Learning vectors: the dot product of vectors General Math 5