- #1
ytht100
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It is from some famous publications. But I seem can get it from rigorous proof after many hours and different methods of trying and Googling.
If we have g as the eigenvector of a symmetric matrix and G is the eigenvalue of the symmetric matrix.
[tex]\left[ {\begin{array}{*{20}{c}} {X11 - G}&{X12}&{X13}\\ {X12}&{X22 - G}&{X23}\\ {X13}&{X23}&{X33 - G} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {g1}\\ {g2}\\ {g3} \end{array}} \right] = 0[/tex]
and:
[tex]\begin{array}{l}
E11 = (X22 - G)(X33 - G) - X_{23}^2\\
E22 = (X11 - G)(X33 - G) - X_{13}^2\\
E33 = (X11 - G)(X22 - G) - X_{12}^2
\end{array}[/tex]
Then:
[tex]g1g1 = E11/(E11 + E22 + E33)?[/tex]
Why?
Thanks a lot!
If we have g as the eigenvector of a symmetric matrix and G is the eigenvalue of the symmetric matrix.
[tex]\left[ {\begin{array}{*{20}{c}} {X11 - G}&{X12}&{X13}\\ {X12}&{X22 - G}&{X23}\\ {X13}&{X23}&{X33 - G} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {g1}\\ {g2}\\ {g3} \end{array}} \right] = 0[/tex]
and:
[tex]\begin{array}{l}
E11 = (X22 - G)(X33 - G) - X_{23}^2\\
E22 = (X11 - G)(X33 - G) - X_{13}^2\\
E33 = (X11 - G)(X22 - G) - X_{12}^2
\end{array}[/tex]
Then:
[tex]g1g1 = E11/(E11 + E22 + E33)?[/tex]
Why?
Thanks a lot!