Timeline for real symmetric matrix has real eigenvalues - elementary proof
Current License: CC BY-SA 3.0
5 events
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| Jan 13, 2013 at 15:38 | comment | added | Chris Godsil | I can do it if I am allowed to use spectral decomposition. Write $A$ as $A_1 + bb^T$, where the first row and column of $A_1$ are both zero. (If needed replace $A$ by $-A$.) Then $$ \det(tI-A) = \det(tI-A_1-bb^T) = \det(tI-A_1)\det(I-(tI-A)^{-1}bb^T) $$ and since $\det(I-uv^T)=1-v^Tu$, we get that $\det(tI-A)/\det(tI-A_1)$ is equal to $1-b^T(tI-A_1)^{-1}b$. Now use spectral decomposition to deduce that the numerators in $b^T(tI-A_1)^{-1}b$ are real. (This argument is logical, but it might not be a lot of fun in a classroom.) | |
| Jan 13, 2013 at 5:54 | comment | added | Brendan McKay | Might it be done using eigenvalue interlacing on the original matrix rather than reducing to tridiagonal form first? | |
| Jan 11, 2013 at 23:33 | vote | accept | marjeta | ||
| Jan 11, 2013 at 23:33 | |||||
| Jan 11, 2013 at 17:30 | history | edited | Igor Khavkine | CC BY-SA 3.0 |
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| Jan 11, 2013 at 16:59 | history | answered | Chris Godsil | CC BY-SA 3.0 |