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Math Help - Why is the sum of 2 positive definite matrices positive definite?

  1. #1
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    Why is the sum of 2 positive definite matrices positive definite?

    Is it because

    If A is pos. def. then z^T*M*z > 0
    If B is pos. def. then z^T*M*z > 0

    so A + B will also be > 0?

    This is the only way I know how to represent a positive definite matrix, so I'm not sure how to start a proof other than something like that?
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  2. #2
    Junior Member nimon's Avatar
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    Hi gummy_ratz,

    All you need to verify is that for any vector x and any two positive definite matrices A and B, then

    x^{T}(A+B)x = x^{T}Ax + x^{T}Bx \geq 0.

    First try and show this for a vector (x_{1},x_{2}) and 2\times 2 matrices

    \left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right) and \left(\begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array} \right)

    Then you could try and do the general case.

    Then do you see how this proves that A+B is positive definite?
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  3. #3
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    Yeah, that makes sense. That was kind of what I was getting at.

    Because x^TAx > 0 since A is pos. def.
    x^TBx > 0 since B is pos. def.

    and so x^T(A + B)x = (x^T*A + x^T*B)x = x^TAx + x^TBx .... both of these are greater than 0 as shown above, so x^TAx + x^TBx > 0.
    And that would be enough?
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by gummy_ratz View Post
    Yeah, that makes sense. That was kind of what I was getting at.

    Because x^TAx > 0 since A is pos. def.
    x^TBx > 0 since B is pos. def.

    and so x^T(A + B)x = (x^T*A + x^T*B)x = x^TAx + x^TBx .... both of these are greater than 0 as shown above, so x^TAx + x^TBx > 0.
    And that would be enough?
    Sure, that's enough: provided it is enough to show that A+B satisfies the definition of being positive definite. So you have to ask yourself: what is the definition of a matrix being positive definite?
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  5. #5
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    Well, as long as the equation produced by xT(A+B)x will always be positive then A + B is positive definite I believe?
    So since xTAx produces a positive equation and xTBx produces a positive equation then xTAx + xTBx will be a positive equation, so A + B is positive definite.
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  6. #6
    Super Member Failure's Avatar
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    Quote Originally Posted by gummy_ratz View Post
    Well, as long as the equation produced by xT(A+B)x will always be positive then A + B is positive definite I believe?
    So since xTAx produces a positive equation and xTBx produces a positive equation then xTAx + xTBx will be a positive equation, so A + B is positive definite.
    A matrix M is defined to be positive definite if for every vector x\neq 0 it is true that x^TM x>0, and you have shown that for M=A+B. (But note that it would not be good enough to show that x^TM x\geq 0 holds.)

    The only problem that I see is that you seem to feel so uncertain about it. Why is that? You seem to know the definition, you seem to have understood the proof, so why don't you just dare trust your own reasoning? - Do dare to trust it.
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