- #1

fluidistic

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## Homework Statement

I must understand the following proof.

Let [itex]A \in \mathbb{R}^{n \times n}[/itex] be a symmetric and positive definite matrix.

Thus there exist a unique factorization of [itex]A[/itex] such that [itex]A=LL^t[/itex] where [itex]L[/itex] is a lower triangular matrix whose diagonal is positive ([itex]l_{ii}>0[/itex])

Demonstration:

[itex]A=A^t[/itex] and [itex]x ^t A x>0[/itex] for [itex]x \neq 0[/itex]. This is the hypothesis on [itex]A[/itex].

It follows that [itex]A[/itex] is invertible, so [itex]A^{-1}[/itex] exist.

Furthermore, considering vectors of the type [itex]x=(x_1, x_2, ..., x_k, 0, ... , 0)[/itex] one can see that the minors of [itex]A[/itex] are also definite positive.

It follows from the [itex]LU[/itex] decomposition theorem that [itex]A[/itex] admits a [itex]LU[/itex] decomposition. In other words I can write [itex]A=LU \Rightarrow A^t=U^t L^t \Rightarrow U^t L^t (L^t)^{-1}=U^t[/itex] and [itex]U(L^t)^{-1}=L^{-1}U^t[/itex].

Where [itex]U(L^t)^{-1}[/itex] is an upper triangular matrix and [itex]L^{-1}U^t[/itex] is a lower triangular matrix. Thus there exist a diagonal matrix [itex]D[/itex] such that [itex]U(L^t)^{-1}=D[/itex].

From this, we have that [itex]U=DL^t[/itex]. Since [itex]A=LU[/itex], [itex]A=LDL^t[/itex]. Therefore [itex]D[/itex] is definite positive so its diagonal elements are positive.

So I can write [itex]A= L'L'^t[/itex] with [itex]L'=L \sqrt D[/itex].

Now in order to complete the proof, I must show that [itex]L'[/itex] is definite positive (I think it will imply that its diagonal entries are all positive). This is where I'm stuck.

I couldn't even show that [itex]\sqrt D[/itex] is positive definite...

Uniqueness proof:

Suppose that there exist [itex]L_1[/itex] such that [itex]A=LL^t=L_1L_1^t[/itex]. Let's show that [itex]L=L_1[/itex].

1)Multiplying by [itex]L^{-1}[/itex] we get [itex]L^t= L^{-1}L_1L_1^t[/itex].

2)Multiplying by [itex](L_1)^{-1}[/itex] we get that [itex]L^t (L_1^t)^{-1}=L_1 ^{-1}L_1L_1^t (L_1^t)^{-1}=L^{-1}L_1=D*[/itex] where [itex]D*[/itex] is diagonal with positive entries.

It follows that:

3)[itex]L_1=LD*[/itex]

4)[itex]L^t =D* L_1 ^t[/itex] I don't understand how the proof reaches this

From 3) and 4), we get (I don't understand the following implication) [itex]D*(LD*)^t=D*D*^t L^t=(D*)^2L^t \Leftrightarrow D*=I \Rightarrow L=L_1[/itex] which complete the proof.

Any help is greatly appreciated.

P.S.:D* is a matrix. * isn't a multiplying sign.