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pippi6

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feixiaolin: ½ð±Ò+1 2014-12-13 20:34:40
ÒýÓûØÌû:
8Â¥: Originally posted by zaq123321 at 2014-12-13 12:45:43
Do you mean my claim is wrong? Or want to add more information?
...

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
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hank612

ÖÁ×ðľ³æ (ÖøÃûдÊÖ)

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ȫȫfly(feixiaolin´ú·¢): ½ð±Ò+3 2014-12-13 15:39:26
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.
We_must_know. We_will_know.
2Â¥2014-12-11 06:28:31
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pippi6

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ÒýÓûØÌû:
2Â¥: Originally posted by hank612 at 2014-12-11 06:28:31
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,  ...

¹ØÓÚµÚÒ»Ì⣬ ÎÒÏëÒ²ÐíÂ¥Ö÷ÕæÕýÏëÎʵÄÊÇ ¡°AB µÄÌØÕ÷Öµ ÊÇ·ñ¾ÍÊÇAµÄÌØÕ÷Öµ³ËÒÔBµÄÌØÕ÷Öµ¡± ÕâÑùÒ»ÀàµÄ¹ØÏµ¡£
3Â¥2014-12-11 07:31:08
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

zaq123321

ר¼Ò¹ËÎÊ (ÖøÃûдÊÖ)

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1. I only know the product of the eigenvalues of AB equals th product of eigenvalues of A and eigenvalues of B.
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4Â¥2014-12-11 10:33:24
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