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Linear Proofs

  1. Sep 4, 2011 #1

    A(rB) = r(AB) =(rA)B where r is a real scalar and A and B are appropriately sized matrices.

    How to even start? A(rbij)=A(rB), but then you can't reassociate...

    Also, a formal proof for Tr(AT)=Tr(A)?

    It doesn't seem like enough to say the diagonal entries are unaffected by transposition..

    Lastly, let A be an mxn matrix with a column consisting entirely of zeros. Show that if B is an nxp matrix, then AB has a row of zeros.

    I can't figure out how to make a proof of this. I know how to say what such and such entry of AB is, but I don't know how to designate an entire column. How do you formally say it will be equal to zero, then...just because the dot product of a zero vector with anything is 0?
  2. jcsd
  3. Sep 5, 2011 #2


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    You don't have to. Every entry in "rB" has a factor of r so every entry in A(rB) has a factor of r so A(rb)= r(AB)= (rA)B

    Why not? Would it be better to say "[itex]A^*_{ij}= A_{ji}[/itex]" so that, replacing j with i, "[itex]A^*_{ii}= A_{ii}[/itex]"? That may look more "formal" but it is really just saying that "the diagonal entries are unaffected by transposition".

    [tex](AB)_{ij}= \sum A_{ik}B_{kj}[/itex]. If the "jth" column of B is all 0s, then the "jth" row of A is all 0.

  4. Sep 5, 2011 #3
    Thank you.



    I can't even see how to begin...
  5. Sep 5, 2011 #4


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    That's a sum of squares!
  6. Sep 5, 2011 #5


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    The definition of rA where r is a real number and A is a matrix is [itex](rA)_{ij}=rA_{ij}[/itex]. The definition of AB where both A and B are matrices is [itex](AB)_{ij}=\sum_k A_{ik}B_{kj}[/itex]. It's not hard to use these definitions to show that the equalities you mentioned are true. Start with [itex](A(rB))_{ij}=\sum_k A_{ik}(rB)_{kj}[/itex].

    All your other questions are also quite easy to answer if you just use these definitions, and the definition of the trace and the transpose: [itex]\operatorname{Tr}A=\sum_i A_{ii},\quad (A^T)_{ij}=A_{ji}[/itex].
    Last edited: Sep 5, 2011
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