anonymous
  • anonymous
Any ideas how to show this, <(s,t),(x,y)> = 2ax + 3ty is or is not an inner product on R^2
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
If you define inner product as: \[<(p,q),(u,v)>=pu+qv\] Then: \[<(s,t),(x,y)>=sx+ty\] If this is equal to the question expression then it is an inner product, or else it is not. So you find the solution to the equation. If you set the vectors to correspond to the answer then it is an inner product. If the vectors are not the set which the equation is valid, then it is not an inner product.
anonymous
  • anonymous
Yea I need to prove the the 4 or 5 properties of inner products
anonymous
  • anonymous
Anyone know if <(s,t),(x,y)> = sy + tx is or is not and inner product on R^2

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anonymous
  • anonymous
I don't think that is the way this question goes. First you try to equate that expression to the actual inner product definition: \[<(s,t),(x,y)> = sx + ty = 2ax+3ty\] Then you solve the equation formed with the second and third part, you can solve for any single variable, I chose y for a good reason though: \[\begin{align}<(s,t),(x,y)> = sx + ty &= 2ax+3ty\\ sx&=2ax+2ty\\ 2ty&=sx-2ax\\ y&=\frac{sx-2ax}{2t}\\ \end{align}\] Now you just answer that for any two vectors (s,t) and (x,y) in which: \[y=\frac{sx-2ax}{2t}\] The expression in the question is an inner product. If y is not that value, than the statement in the question is not an inner product.
anonymous
  • anonymous
wrote it wrong the a is an s but thanks got it

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