Making a Set a Basis: Is V Spanned?

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SUMMARY

The set S={cos(x)^2, sin(x)^2, cos(2x)} does not form a basis for the vector space V due to the linear dependence of its elements, specifically that cos(2x) can be expressed as a linear combination of cos(x)^2 and sin(x)^2. The equation a(cos(x)^2) + b(sin(x)^2) = 0 must be evaluated to determine if the first two vectors are linearly independent. Evaluating this equation at x=0 and x=π/2 provides insight into the linear independence of cos(x)^2 and sin(x)^2.

PREREQUISITES
  • Understanding of linear combinations in vector spaces
  • Familiarity with trigonometric identities, specifically cos(2x) = cos^2(x) - sin^2(x)
  • Knowledge of linear independence and dependence concepts
  • Basic calculus skills for evaluating trigonometric functions
NEXT STEPS
  • Study linear independence and dependence in vector spaces
  • Learn about trigonometric identities and their applications in linear algebra
  • Explore methods for evaluating linear combinations of functions
  • Investigate the implications of evaluating functions at specific points, such as x=0 and x=π/2
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Students and educators in mathematics, particularly those studying linear algebra and trigonometry, as well as anyone interested in understanding the concepts of vector spaces and linear independence.

torquerotates
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Ok, so I have the set S={cosx^2, sinx^2, cos(2x)} that spans V.

So obviously this is not a basis. Because the third is a linear combination of the first; ie. cosx^2-sinx^2=cos(2x). But if I were to take the first 2 vectors, would they span V? That is does the equation a(cosx^2)+b(sinx^2)=0 imply a=0& b=0?
 
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torquerotates said:
Ok, so I have the set S={cosx^2, sinx^2, cos(2x)} that spans V.

So obviously this is not a basis. Because the third is a linear combination of the first; ie. cosx^2-sinx^2=cos(2x). But if I were to take the first 2 vectors, would they span V? That is does the equation a(cosx^2)+b(sinx^2)=0 imply a=0& b=0?

Do you mean: Are the first two vectors linearly independent?

Use [itex]\cos^2x = 1 - \sin^2x[/itex].
 
torquerotates said:
does the equation a(cosx^2)+b(sinx^2)=0 imply a=0& b=0?

try evaluating at x=0 and x=[itex]\pi/2[/itex]
 

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