PROOF: Independent vectors and spanning vectors

In summary, in order to span a space of dimension m, we need at least m vectors which are independent and form a basis for that dimension. To determine if vectors are linearly independent, we can use the condition of au+bv=0 and perform row reduction on the stacked matrix. If there is no row of zeros, then the vectors are independent.
  • #1
Ella087
3
0
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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  • #2
do you know how to do proofs?
 
  • #3
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".
 
  • #4
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.
 
  • #5
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.
 
  • #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.
 
  • #7
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:
  • #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
 
  • #9
tomyus said:
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...

transgalactic said:
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.
 

1. What is the difference between independent vectors and spanning vectors?

Independent vectors are a set of vectors that cannot be written as a linear combination of each other. Spanning vectors, on the other hand, are a set of vectors that can create any vector in a given vector space through linear combinations.

2. How do you determine if a set of vectors is independent?

A set of vectors is independent if the only solution to the equation c1v1 + c2v2 + ... + cnvn = 0 is when all the coefficients (c1, c2, ..., cn) are equal to zero.

3. Can a set of independent vectors also be spanning?

Yes, a set of independent vectors can also be spanning. If a set of vectors is both independent and spanning, it is called a basis for the vector space.

4. How do you find the dimension of a vector space?

The dimension of a vector space is equal to the number of vectors in a basis for that vector space. For example, if a basis for a vector space has 3 vectors, then the dimension of that vector space is 3.

5. Is every set of vectors in a vector space spanning?

No, not every set of vectors in a vector space is spanning. In order for a set of vectors to be spanning, they must be able to create any vector in that vector space through linear combinations. If a set of vectors is not spanning, then it is not a basis for that vector space.

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