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So the big theory is every hilbert space operator is a linear combination of projections, and for matrices, Any Hermitian Matrix is the Linear Combination of Four Projections.

But the proof in the paper does not give a explicit algorithm to compute those four projections, thus I am curious how we should look for those four projections.

Thanks!

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    Do you really think $[0,0;1,0]$ is Hermitian?2012-07-07
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    @did I am asking about general matrices, which are linear combinations of hermitian matrices.2012-07-07
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    @did Since any matrix is the sum of its real part and imaginary part, both of which are Hermitian. Such a result for Hermitian matrices would automatically apply to general matrices.2012-07-07
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    @did By real part he means $\frac{A+A^*}2$.2012-07-07
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    @azarel: Thanks.2012-07-07
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    A self-adjoint $n$-by-$n$ matrix is a linear combination of at most $n$ *commuting* projections, which is probably more useful. Normal matrices can be characterized as linear combinations of commuting projections.2012-07-08
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    @JonasMeyer I have not learnt about this. Can you give some reference maybe? Also, I still hope for an explicit algorithm for finding these projections. Thanks!2012-07-08
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    @HuiYu: This is a consequence of the spectral theorem for normal matrices: http://en.wikipedia.org/wiki/Spectral_theorem#Normal_matrices. You can check directly that a linear combination of commuting self-adjoint idempotents is normal. Conversely, the spectral theorem says that a normal operator has an orthogonal basis of eigenvectors, so the projections in question can be taken to be the othogonal projections onto the eigenspaces. (This has probably zero bearing on your question.)2012-07-08

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