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I have this equation:$$D=LRL^{H}$$ where D is diagonal matrix, L is lower triangular matrix, R is positive definite matrix. How can one obtain these equations from above equation?$$R^{-1}=L^{H}D^{-1}L$$ $$R^{-1}=(D^{\frac{-1}{2}}L)^{H}(D^{\frac{-1}{2}}L)$$

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What is $H$? in the equations? –  utdiscant Apr 16 '12 at 7:03
    
H is hermitian form of matrix –  hoka Apr 16 '12 at 7:08
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Not hermitian form: hermitian transpose, also called conjugate transpose, Hermitian conjugate, etc. –  Robert Israel Apr 16 '12 at 7:10
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Can you use $(AB)^{-1}=B^{-1}A^{-1}$ and $(AB)^H=B^HA^H$? I should think your formulas would follow from those. –  Gerry Myerson Apr 16 '12 at 7:10
    
yes, I can use these properties, but how can I apply them? –  hoka Apr 16 '12 at 7:16

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Assuming $L$ is invertible, so is $L^H$. Take the inverse of both sides of your equation, noting that the inverse of a product of invertible matrices is the product of the inverses in reverse order:

$$ D^{-1} = (L R L^H)^{-1} = (L^H)^{-1} R^{-1} L^{-1}$$

Now can you see how to get $L^H D^{-1} L$ on the left side?

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Thank you! Could you please introduce me a good reference for these properties of matrix? –  hoka Apr 16 '12 at 7:31
    
@hoka: any good linear algebra book ought to do you well. –  J. M. Apr 17 '12 at 2:18

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