a) Find the Ker (T)
b) Find the Image (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.
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.