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Det(A+B) ? det(A) + det(B)

  1. Sep 14, 2011 #1
    det(A+B) ?? det(A) + det(B)

    I'm guessing greater than but I'm not too sure. I need a proof on this so I can be assured of it and then use the statement to prove something else.

    any hint (or link to proof) would be much appreciated.

    edit:

    x*det(A) ?? det(x*A)
    what's the relation there?
     
  2. jcsd
  3. Sep 14, 2011 #2

    micromass

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    Re: det(A+B) ?? det(A) + det(B)

    Neither holds, take

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

    for one equality and


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

    for the other.

    For the last question, we do have the equality

    [tex]det(xA)=x^n detA[/tex]

    where A is the rank of the matrix. From this we can deduce that neither of the inequalities between det(xA) and xdet(A) holds true. Just take x>1 and x<1.
     
  4. Sep 14, 2011 #3
    Re: det(A+B) ?? det(A) + det(B)

    thanks. i better find an alternative way to my proof ;s
     
  5. Sep 14, 2011 #4

    AlephZero

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    Re: det(A+B) ?? det(A) + det(B)

    Another simple counter-example is
    [tex]A = \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \qquad
    B = \begin{pmatrix} 0 & 0 \\ 0 & b \end{pmatrix}[/tex]

    det(A) = det(B) = 0, det(A+B) = ab
     
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