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- #1

- Apr 13, 2013

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I want to show that the lower triangular matrix L,which has the identity $A=LL^{T}$ ,where A is a positive-definite and symmetric matrix,is unique.

That's what I have done so far:

Suppose that there are two matrices,with that identity.

Then, $$A=LL^{T}=MM^{T} \Rightarrow M^{-1}L=M^{T}(L^{T})^{-1}$$

$M^{-1}L$ is a lower triangular matrix and $M^{T}(L^{T})^{-1}$ is an upper triangular matrix,so,because of the fact that they are equal,they must be equal to the diagonal matrix.

So, $$L=MD \text{ and } M^{T}=DL^{T} $$

$$L=MD \Rightarrow L^{T}=D^{T}M^{T}$$

Replacing it at the other relation we have,we get $$M^{T}=DD^{T}M^{T}$$

So,it must be $DD^{T}=I \Rightarrow DD=I \Rightarrow D=I$.

Is this right?