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anonymous
 4 years ago
Prove that set of all fourth degree polynomials is not a vector space
anonymous
 4 years ago
Prove that set of all fourth degree polynomials is not a vector space

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Weelll, it is not so simple to prove a false assertion....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since it is closed to linear combinations and closed to multiplication by a scalar  it IS a vector space. 4Dimensional by the way.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry  5 dimensional

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cud u plz tell me the exact steps to write :) :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know what your Lecturer/book THINKS ( and he/she is wrong !)  it thinks that P(x) = 7x +3 is not 4th degree. But unfortunately for him/her it formally IS

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Consider 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
 4 years ago
Best ResponseYou've already chosen the best response.0Adding 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
 4 years ago
Best ResponseYou've already chosen the best response.0The fact that sometimes a_4 + b_4 = 0 shall not be of any problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks @Priyanka12081
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