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Kernal and Range of a Linear Transformation 
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#1
Apr2111, 04:38 PM

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Let L:p2 >>> p3 be the linear transformation defined by L(p(t)) = t^2 p'(t).
(a) Find a basis for and the dimension of ker(L). (b) Find a basis for and the dimension of range(L). The hint that I get is to begin by finding an explicit formula for L by determining L(at^2 + bt + c). Does this hint mean let p(t) = at^2 + bt + c? Then, I find that t^2 p'(t) = 2at^3 + bt^2. Then, I conclude that the basis for ker(L) = {1}. Is it right? Also, how to find range(L)? Thanks 


#2
Apr2111, 05:58 PM

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Factorise...
[tex] L(at^{2}+bt+c)=2at^{3}+bt^{2}=t^{2}(2at+b) [/tex] This will be zero when t=0 or t=b/2a, so... 


#3
Apr2111, 06:20 PM

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#4
Apr2111, 09:59 PM

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Kernal and Range of a Linear Transformation
is the basis for ker(L) {t, 1} and the basis for range(L) {t^3, t^2}?



#5
Apr2111, 10:08 PM

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#6
Apr2111, 10:12 PM

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#7
Apr2211, 06:58 AM

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Since L(at^2 + bt + c) = 2at^3 + bt^2
No matter what value of t and 1, 2a^3 + bt^2 should always give me 0 vector. So, I have a question why Ker(L) does not have t as a basis? Another question is dim Ker(L) + dim range(L) = dim (p3) by thm. since the dim range(L) = 2 and dim (p3) = 4, why dim ker(L) not equal to 2? 


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Apr2211, 07:12 AM

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#9
Apr2211, 08:32 AM

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