# Vector Space

• Apr 19th 2008, 04:30 AM
pearlyc
Vector Space
Are the vectors 6, 3 sin^2x and 2cos^2x in F(-infinity,infinity) the space of all functions from R -> R linearly dependent or linearly independent?
• Apr 19th 2008, 04:39 AM
Plato
Consider: $\displaystyle 6 + \left( { - 2} \right)\left[ {3\sin ^2 (x)} \right] + \left( { - 3} \right)\left[ {2\cos ^2 (x)} \right]$
• Apr 19th 2008, 04:42 AM
pearlyc
Thanks for the help, I kinda get the approach but I don't quite know how to present it. Like, where do I start?

As I know, the way to prove that something is linearly independent or not is through row reduction then compare the ranks. What about this one?
• Apr 19th 2008, 04:51 AM
Plato
QUOTE=pearlyc;132175]As I know, the way to prove that something is linearly independent or not is through row reduction then compare the ranks. What about this one?[/QUOTE]
You must first learn to general definitions.
If one can find a nontrivial linear combination that equals the zero then the collection is dependent. Row reductions are applied to matrices and not to general spaces.
• Apr 19th 2008, 04:54 AM
pearlyc
Is it alright if I were to approach the question like this,

A linear dependent vector is when we have n vectors and one can be expressed as a linear combination of the rest. When this is impossible, we say the vectors are linearly indepedent.

Considering,

a(3sin^2 x) + b(2cos^2 x) = 6
a(3sin^2 x) + b(2cos^2 x) = 6(sin^2 x + cos^2 x)
3asin^2 x + 2bcos^2 x = 6sin^2 x + 6cos^2 x

Comparing coefficients,

3a = 6
Therefore, a = 2

2b = 6
Therefore, b = 3

Since the vectors could be expressed as a linear combination, therefore we can say that the vectors are linearly dependent.

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Do you think this working makes sense and would be acceptable?
• Apr 19th 2008, 05:01 AM
Plato
Quote:

Originally Posted by pearlyc
Do you think this working makes sense and would be acceptable?

Yes.