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wxbddp

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hwzxaww

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2Â¥2013-07-29 20:10:28
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hwzxaww

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3Â¥2013-07-29 20:20:17
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wxbddp

гæ (СÓÐÃûÆø)


ÒýÓûØÌû:
2Â¥: Originally posted by hwzxaww at 2013-07-29 20:10:28
ÌØÕ÷Öµ²»±ä£¬ÌØÕ÷ÏòÁ¿¸Ä±ä¡£

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4Â¥2013-07-29 20:38:59
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Halmos

Ìú³æ (СÓÐÃûÆø)


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The matrices $AB$ and $BA$ are similar provided either $A$ or $B$ is invertible.

If both $A$ and $B$ are singular (and square), a limiting argument involving $A +\epsilon I$
is useful. In this case $AB$ and $BA$ still have the same eigenvalues with the same
multiplicities. However, in general,  $AB$ is not similar to $BA$, . Their Jordan forms may be different, in the sizes of the blocks associated with the eigenvalue $0$.
6Â¥2013-07-30 00:09:39
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wxbddp

гæ (СÓÐÃûÆø)


ÒýÓûØÌû:
6Â¥: Originally posted by Halmos at 2013-07-30 00:09:39
The matrices $AB$ and $BA$ are similar provided either $A$ or $B$ is invertible.

If both $A$ and $B$ are singular (and square), a limiting argument involving $A +\epsilon I$
is useful. In this c ...

What you write is not quite an answer to my question............
7Â¥2013-07-30 08:31:04
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

yangrui123

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8Â¥2013-07-30 12:43:01
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Halmos

Ìú³æ (СÓÐÃûÆø)


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ÒýÓûØÌû:
7Â¥: Originally posted by wxbddp at 2013-07-30 08:31:04
What you write is not quite an answer to my question...............

Sorry, I went too far... but from which it's very clear the answer to your question.
9Â¥2013-07-31 06:44:04
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lxc_422

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10Â¥2013-10-08 15:49:13
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lincoln24

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11Â¥2014-04-03 17:09:52
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wxbddp

гæ (СÓÐÃûÆø)


ÒýÓûØÌû:
11Â¥: Originally posted by lincoln24 at 2014-04-03 17:09:52
»°ËµÕâ¸öÎÊÌâÓнáÂÛÁËÂ𡣡£¡£¡£

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12Â¥2014-04-03 18:10:53
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helloqiting

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