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Is this in general true (about projection matrices)?

  1. Sep 9, 2015 #1

    td21

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    $$A$$ is a hermitian matrix with eigenvalues +1 and -1. Let $$\left|+\right>$$ and $$\left|-\right>$$ be the eigenvector of $$A$$ with respect to eigenvalue +1 and eigenvalue -1 respectively.
    Therefore, $$P_{+} = \left|+\right>\left<+\right|$$ is the projection matrix with respect to eigenvalue +1. $$P_{-} = \left|-\right>\left<-\right|$$ is the projection matrix with respect to eigenvalue -1.

    We all know that $$A = P_{+} + (-1)P_{-}$$. But is $$I = P_{+} + P_{-}$$ true? $$I$$ is the identity matrix.
     
    Last edited: Sep 9, 2015
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  3. Sep 9, 2015 #2

    DEvens

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    It is the identity matrix in context of the vector space spanned by ## \left|+\right>## and ## \left|-\right>##.

    Why do I weasel? Because you can probably find vectors that are orthogonal to ## \left|+\right>## and ## \left|-\right>##, and it would be surprising if you applied the identity matrix to a vector and got zero. So ##P_+ + P_-## is only the identity matrix with respect to the appropriate vector space.

    How do you prove it? You apply ##I## to an arbitrary combination of ##x\left|+\right> + y\left|-\right>## and see what you get. And you determine what is the square of ##I##.
     
  4. Sep 9, 2015 #3

    Geofleur

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    If the eigenvectors ## | + \rangle ## and ## | - \rangle ## form a basis of the underlying vector space, then yes. Because then, for any ## | v \rangle ##,

    ## (P_+ + P_-) | v \rangle = | + \rangle \langle + | v \rangle + | - \rangle \langle - | v \rangle = I | v \rangle ##.

    That is, if we can expand any ket in terms of the eigenvectors, then we can write the identity as a sum of the corresponding projection operators.

    Oops, DEvens beat me to it!
     
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