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## Main Question or Discussion Point

Suppose A is a non-singular matrix. A

S

(S

(AS)

So, A

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

A

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

Therefore:

A

So, A

Is my proof right?

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

^{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}=ΛSo, 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!