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Homework Help: Trace prove question

  1. Jun 21, 2011 #1
    i need to prove that if tr(A^2)=0

    then A=0



    we have a multiplication of 2 the same simmetrical matrices

    why there multiplication is this sum formula

    [iTEX]

    A*A=\sum_{k=1}^{n}a_{ik}a_{kj}

    [/iTEX]



    i know that wjen we multiply two matrices then in our result matrix

    each aij member is dot product of i row and j column.

    dont understand the above formula.





    and i dont understand how they got the following formula:

    then when we calculate the trace (the sum of the diagonal members)

    we get

    [TEX]

    tr(A^{2})=\sum_{i=1}^{n}A_{ii}^{2}=\sum_{i=1}^{n}(\sum_{i=1}^{n}a_{ik}a_{ki})

    [/TEX]

    and because the matrix is simmetric then the trace is zero

    why?


    i need to prove that if tr(A^2)=0



    then A=0



    can you explain the sigma work in order to prove it?
     
  2. jcsd
  3. Jun 21, 2011 #2
    Hi nhrock3! :smile:

    This is false. Consider

    [tex]A=\left(\begin{array}{cc} 0 & 1\\ 0 & 0\\ \end{array}\right)[/tex]

    then tr(A2)=0, but A is not zero.

    What does the exercise say precisely?
     
  4. Jun 21, 2011 #3
    if A is a simetric matrix and if tr(A^2)=0 then A=0
     
  5. Jun 21, 2011 #4
    Ah, you should have said they were symmetric! :smile:

    If

    [tex]A=\left(\begin{array}{ccccc}
    a_{11} & a_{12} & a_{13} & ... & a_{1n}\\
    a_{12} & a {22} & a_{23} & ... & a_{2n}\\
    a_{13} & a_{23} & a_{33} & ... & a_{3n}\\
    \vdots & \vdots & \vdots & \ddots & \vdots\\
    a_{1n} & a_{2n} & a_{3n} & ... & a_{nn}\\
    \end{array}\right)[/tex]

    then what will be the diagonal of A2?? In particular, can you show that the diagonal contains only positive values?
     
  6. Jun 21, 2011 #5
    each member of the diagonal on the A^2 matrix its number of row equals the column number

    so the second member in the diagonal is the dot product of the second row with the second column
    etc..
    so i get this expression
    [tex]tr(A^{2})=\sum_{k=1}^{n}\sum_{i=1}^{n}a_{ki}a_{ik}[/tex]
    so because its simetric and equlas zero
    [tex]tr(A^{2})=\sum_{k=1}^{n}\sum_{i=1}^{n}(a_{ki})^2=0[/tex]
    so we get a sum of squeres and in order for them to be zero
    then each one of them has to be zero
    thannkkks :)
     
    Last edited: Jun 21, 2011
  7. Jun 21, 2011 #6
    Indeed, so the k'th member in the diagonal is

    [tex]\|(a_{1k},...,a_{nk})\|^2[/tex]

    Do you see that?
     
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