- #1
BiBByLin
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Suppose A is a non-singular matrix. AT is the transpose matrix of A. Therefore, the eigenvalue of A can be expressed as:
S-1AS=Λ
(S-1AS)T=ΛT
(AS)TS-T=STATS-T=Λ
So, AT and A share the same eigenvalue and eigenvector.
Here, x is the base eigenvector of A. Hence, span{x} is the eigenvector of A.
ATAx=ATΛx.
We consider the new eigenvector Λx as the eigenvector of A or AT.
Therefore:
ATAx=ATΛx=ΛΛx.
So, ATA is symmetric positive definite.
Is my proof right?
It is very urgent for my program, pls help me. Thanks a lot!
S-1AS=Λ
(S-1AS)T=ΛT
(AS)TS-T=STATS-T=Λ
So, AT and A share the same eigenvalue and eigenvector.
Here, x is the base eigenvector of A. Hence, span{x} is the eigenvector of A.
ATAx=ATΛx.
We consider the new eigenvector Λx as the eigenvector of A or AT.
Therefore:
ATAx=ATΛx=ΛΛx.
So, ATA is symmetric positive definite.
Is my proof right?
It is very urgent for my program, pls help me. Thanks a lot!