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Consider the function $T: P_1 \rightarrow P_1$ defined by $T(at+b) = (3a+b)t + 4a +2b$.
Show that T is a linear transformation.

Solution: From the definition of a linear transformation we must show that the function satisfy the two properties of a linear transformation(vector addition and scalar multiplication)
This is what I have:
$T((at+b)+(a_1t+b_1)) = T((a+a_1)t + (b+b_1)) = T(at+b)+T(a_1t+b_1)$

But I don't think this shows that the function holds for vector addition.

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Does your proof still work if I define $T(at+b)=1$? – wj32 Nov 24 '12 at 9:38
up vote 2 down vote accepted

Be careful, your proof does not prove much! We have to show that $T((at+b)+(a_1t+b_1)) = T(at+b)+T(a_1t+b_1)$ and $ T( \alpha (at+b))= \alpha T(at+b)$ for any $\alpha$ in the field (over which the polynomial are defined). $T((at+b)+(a_1t+b_1))=T((a+a_1)t+(b+b_1))=(3(a+a_1)+(b+b_1))t+4(a+a_1)+2(b+b_1)=(3a+3a_1+b+b_1)t+4a+4a_1+2b+2b_1=(3a+b)t+2a+2b+(3a_1+b_1)t+4a_1+2b_1=T(at+b)+T(a_1t+b_1)$

and also we have: $T( \alpha (at+b))=T(\alpha at +\alpha b)=(3(\alpha a)+\alpha b)t+4(\alpha a)+2(\alpha b)=\alpha(3a+b)t+\alpha (4a)+\alpha (2b)=\alpha T(at+b)$

(Of course, I used the commutative, associative and distributive laws for a field).

So $T$ is indeed a linear transformation.

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Thank You so much for the explanation and for editing my question. – Diana Martinez Nov 25 '12 at 1:45

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