L(x) = Ax = (3 -4 k 0) (x_1 x_2) Prove that L is subjective if and only if k not

Question

L(x) = Ax = (3 -4 k 0) (x_1 x_2) Prove that L is subjective if and only if k notequalto 0. Assume that V is a vector space over R, with V notequalto {0}, and L: V rightarrow V is a linear map such that L(L(v)) = 0 for all v elemntof V. Prove that L is not injective. Let L: R_3 rightarrow P_2 be a linear function. Prove that L is injective if and only if the following two sets are equal: {(x, y, z) elemntof R_3: L((x,y,z)) = 0} = {(0, 0, 0)}. Note, in the first occurrence in the previous equation, 0 is the polynomial, that is the function, z: R rightarrow R, such that z(t) = 0 for all t elementof R. It is, of course, also the polynomial whose coefficients are all zero. Here are some steps: 1. recognize that this is of the from A doubleheadarrow B, and so you must provide a separate proof for each of A rightarrow B and b rightarrow A. 2. it could be helpful to try to do an argument by contra positive to establish that A rightarrow B is true. 3. when working on B rightarrow A, suppose you are investigating the possibility that x, y elementof R_3 and L(x) = L(y). What does the fact that L is linear say about L(x) - L(y)?

Explanation / Answer

For L to be injective , ker(L)={0}

Since, V is non zero so there is some v in V so that v is not 0

Since L is injective, w=L(v) is non zero

Hence, L(w) is also non zero again because L is injective

But, L(w)=L(L(v))=0

So a contradiction

Hence, L is not injective

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