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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
Not hermitian form: hermitian transpose, also called conjugate transpose, Hermitian conjugate, etc. – Robert Israel Apr 16 '12 at 7:10
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
up vote 1 down vote accepted

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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