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Question: given $T: P_2 →P_3$ be linear transformation given by,

$T(p(x))= (x+1)p(x)$

Find the range(T) and basis for it.

My attempt: since $B =\{1, x, x^2\}$ form basis for $P_2$

So $Range(T)= span\{T(1), T(x), T(x^2)\}$

$= span\{x+1, x^2+ x , x^3 + x^2\}$

$ = \{a(x+1)+ b(x^2 +x)+ c(x^3 + x^2) |a, b, c ∈R \}$

and hence

$Basis$ for $range(T) = \{x+1, x^2+ x, x^3+ x^2\}$

Is my attempt/answer is correct? (but it does not matches with key. May be key is wrong!)

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  • $\begingroup$ Your reasoning let you say that $\{x+1,x^2+x,x^3+x^2\}$ is a system of generators for range(T). A basis is a system of generators which are linearly independent as well. $\endgroup$
    – Federico
    Aug 26, 2017 at 13:24
  • $\begingroup$ So isn't the above basis is linearly independent? I think it is linearly independent. Is am i correct? $\endgroup$
    – Akash Patalwanshi
    Aug 26, 2017 at 13:28
  • $\begingroup$ They are linearly independent, I meant that you should show it to complete the solution. $\endgroup$
    – Federico
    Aug 26, 2017 at 13:46
  • $\begingroup$ Ya thank u. So is my answer is correct? Means the basis that I find is correct? Because in the key the basis is not this :-( $\endgroup$
    – Akash Patalwanshi
    Aug 26, 2017 at 13:56

1 Answer 1

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Your basis is indeed a basis for the range of T. The reason why your answer might not be the same as another answer is because there are infinitely many basis for a vector space. This is because we can just change the co-efficients of the vectors in our basis or add two vectors together and keep the result at the expense of an old one. As long as you arrive at the same answer (there exists/doesn't exist a basis) as your key, then you are correct. Without any context no specific basis is more beautiful or better than any other basis.

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