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^{T}is the transpose matrix of A. Therefore, the eigenvalue of A can be expressed as:

S

^{-1}AS=Λ

(S

^{-1}AS)

^{T}=Λ

^{T}

(AS)

^{T}S

^{-T}=S

^{T}A

^{T}S

^{-T}=Λ

So, A

^{T}and A share the same eigenvalue and eigenvector.

Here, x is the base eigenvector of A. Hence, span{x} is the eigenvector of A.

A

^{T}Ax=A

^{T}Λx.

We consider the new eigenvector Λx as the eigenvector of A or A

^{T}.

Therefore:

A

^{T}Ax=A

^{T}Λx=ΛΛx.

So, A

^{T}A is symmetric positive definite.

Is my proof right?

It is very urgent for my program, pls help me. Thanks a lot!