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?$$R^{-1}=L^{H}D^{-1}L$$ $$R^{-1}=(D^{\frac{-1}{2}}L)^{H}(D^{\frac{-1}{2}}L)$$
How can I do this transformation?
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matrices
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1What is $H$? in the equations? – 2012-04-16
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0H is hermitian form of matrix – 2012-04-16
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1Not hermitian form: hermitian transpose, also called conjugate transpose, Hermitian conjugate, etc. – 2012-04-16
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1Can you use $(AB)^{-1}=B^{-1}A^{-1}$ and $(AB)^H=B^HA^H$? I should think your formulas would follow from those. – 2012-04-16
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0yes, I can use these properties, but how can I apply them? – 2012-04-16
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0Also you need $L$ to be invertible (which is not automatically true) in order for $D^{-1}$ to exist. If it is invertible, $D$ is positive definite and $D^{-1/2}$ exists (just take the $-1/2$ power of each diagonal element). – 2012-04-16
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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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0Thank you! Could you please introduce me a good reference for these properties of matrix? – 2012-04-16
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0@hoka: any good linear algebra book ought to do you well. – 2012-04-17