anonymous
  • anonymous
Prove that set of all fourth degree polynomials is not a vector space
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Weelll, it is not so simple to prove a false assertion....
anonymous
  • anonymous
Since it is closed to linear combinations and closed to multiplication by a scalar - it IS a vector space. 4-Dimensional by the way.
anonymous
  • anonymous
Sorry - 5 dimensional

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anonymous
  • anonymous
cud u plz tell me the exact steps to write :) :)
anonymous
  • anonymous
I know what your Lecturer/book THINKS ( and he/she is wrong !) - it thinks that P(x) = 7x +3 is not 4-th degree. But unfortunately for him/her it formally IS
anonymous
  • anonymous
Consider the set of all expressions of the form\[\sum_{i=0}^{4} a_i * x^i\] where a_i are real/complex and x is a formal letter.
anonymous
  • anonymous
Adding the Linear combination of two objects like that will always result in the same FORM\[\sum_{i=1}^{4} a_i x^i + \sum_{i=1}^{4} b_i x^i = \sum_{i=1}^{4} (a_i + b_i) x^i\]
anonymous
  • anonymous
The fact that sometimes a_4 + b_4 = 0 shall not be of any problem
anonymous
  • anonymous
Thanks :)
anonymous
  • anonymous
Thanks @Priyanka12081

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