can anyone show steps? thank you so much! Let L : V rightarrow W be a linear tra
ID: 1888451 • Letter: C
Question
can anyone show steps? thank you so much!
Let L : V rightarrow W be a linear transformation. Show that ker L = {0V} if and only if L is one-to-one: First, suppose that kerL = {0V}. Show that L is one-to-one. Think about methods of proof-does a proof by contradiction, a proof by induction, or a direct proof seem most appropriate? Now, suppose that L is one-to-one. Show that ker L = {0V}. That is, show that 0V is in ker L, and then show that there are no other vectors in ker L.Explanation / Answer
Suppose that kerL = {0_V}. Assume L(x) = L(y) for some x,y in V. Then L(x)L(y)^(-1) = 0_W, where e is the identity of W. Then L(x)L(y^-1) = e implying L(xy^-1) = e. That means xy^-1 is in the kerL. But kerL = {0_V} so xy^-1 = 0_V or x = y. Therefore L is one-to-one. (Note: The best way to prove f : X -> Y is one to one is to assume f(x)=f(y) and show x=y). Now, suppose L is one-to-one. Clearly L(0_V) = 0_W (should be easy to prove on your own since L is a linear transformation) so we know 0_V is in kerL. Now suppose x is also in kerL. Then L(x) = 0_W. But if both L(x) = 0_W and L(0_V) = 0_W, then x = 0_V since L is one-to-one. Therefore there are no other elements in kerL other than 0_V. QED :)
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