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pippi6
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feixiaolin: ½ð±Ò+1 2014-12-13 20:34:40
feixiaolin: ½ð±Ò+1 2014-12-13 20:34:40
|
Well, I am afraid so, if your do claim that product of the eigenvalues of AB equals th product of eigenvalues of A and eigenvalues of B. It is clear from the example given by hank612 at floor 5. There both A and B have two eigenvalues a and 1/a, but AB has eigenvalues x and 1/x, which are clearly not products of a , 1/a or whatsoever. I must confest that was my initial guess as well but I didn't dare to speak out. |
9Â¥2014-12-13 17:20:14
hank612
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ȫȫfly(feixiaolin´ú·¢): ½ð±Ò+3 2014-12-13 15:39:26
ȫȫfly(feixiaolin´ú·¢): ½ð±Ò+3 2014-12-13 15:39:26
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1. ÈÎÈ¡¿ÉÄæ¾ØÕóA, ÄÇôI µÄÌØÕ÷ÖµºÍ A¼°A^{-1} µÄÌØÕ÷Öµ ºÁÎÞ¹ØÏµ. 2. I +DM µÄÌØÕ÷Öµ µ±È»ÊÇ 1+(DMÌØÕ÷Öµ) ¿ÉÊÇÓÉÓÚËùÓеľØÕóA¶¼ÔÚ¿ÉÒÔÕÒµ½Ò»¸öÓϾØÕóU, ʹµÄ U^T * A * U = D * M ÆäÖÐD ÊǶԽÇÕó, M ÊǶԳÆÕó, ËùÒÔ ÒÀÈ»ÊÇ ºÁÎÞ¹ØÏµ Ìõ¼þ"diag(t)ËùÓжԽÇÔªËØÊÇÕýµÄ¶Ô½ÇÕó¡£MÊÇÕý¶¨¾ØÕó" »ù±¾ÉÏûÓÃ. |

2Â¥2014-12-11 06:28:31
pippi6
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3Â¥2014-12-11 07:31:08
zaq123321
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4Â¥2014-12-11 10:33:24














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