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Math Help - Using the rank-nullity theorem in a proof

  1. #1
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    Using the rank-nullity theorem in a proof

    I'm stuck on this proof, can anyone help?

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  2. #2
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    Quote Originally Posted by kevinlightman View Post
    I'm stuck on this proof, can anyone help?

    The first is pretty straight forward. In general, to prove " A\subset B" you show: "if y is a member of A then y is a member or B".

    Here you want to prove: Im(S+ T)\subset Im(S)+ Im(T)
    If y is in Im(S+ T) then y= (S+ T)x for some x in V. But "(S+ T)x" is defined as Sx+ Tx so we have y= Sx+ Tx. Let y_1= Sx and y_2= Tx. Then y_1 is in Im(S), y_2 is in Im(T), and y= y_1+ y_2. Therefore, y is in Im(S)+ Im(T). The rest of that should be straight forward.

    The second may be even simpler. Saying "ST= 0" mean (ST)v= 0 for all v. But that is the same as S(Tv)= 0 so that Tv is in the kernel of S for all v.
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