Let T: P2 ----> P2 be defined by T(a2x^2 + a1(x) + a0) = (a2+a1)x^2 +a1x + 2a0
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Let T: P2 ----> P2 be defined by T(a2x^2 + a1(x) + a0) = (a2+a1)x^2 +a1x + 2a0

[From: ] [author: ] [Date: 12-07-02] [Hit: ]
the image must be all of IP². Alternatively,is any polynomial in IP²,a_0 = ½ p_0,T(a_2 x² + a_1 x + a_0) = (p_2 - p_1 + p_1) x² + p_1 x + 2(½ p_0) = p_2 x² + p_1 x + p_0 = p(x).This shows that any polynomial in IP² is in the image of T.......
a) Find the Ker (T)

b) Find the Image (T)

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a) Suppose T(a_2x² + a_1x + a_0) = 0---so that the input is in the kernel---then

2a_0 = 0, a_1 = 0, and a_2 + a_1 = 0.

These three equations have solution a_0 = a_1 = a_2 = 0; and this is the only solution. So the kernel contains only the zero polynomial

Ker(T) = {0}.

b) Because the kernel contains only the zero polynomial, the image must be all of IP². Alternatively, we can show that the image is all of IP² by noting that if

p(x) = p_2 x² + p_1 x + p_0

is any polynomial in IP², then setting

a_0 = ½ p_0, a_1 = p_1 and a_2 = p_2 - p_1

gives

T(a_2 x² + a_1 x + a_0) = (p_2 - p_1 + p_1) x² + p_1 x + 2(½ p_0) = p_2 x² + p_1 x + p_0 = p(x).

This shows that any polynomial in IP² is in the image of T.

---To find the formulas for a_0, a_1, a_2, just solve the system a_2 + a_1 = p_2, a_1 = p_1, 2a_0 = p_0.
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