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pippi6

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19楼: Originally posted by zaq123321 at 2014-12-14 23:14:56
1. On spectral radius, we have r(AB)<= r(A)r(B).

With results of floor 5, assuming x>1, a>1, then we have  r(AB)=x, and  r(A)=r(B)=a,
how can you deduct that x <=a^2, when a and x are arbitary?
21楼2014-12-15 09:03:40
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zaq123321

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21楼: Originally posted by pippi6 at 2014-12-15 09:03:40
With results of floor 5, assuming x>1, a>1, then we have  r(AB)=x, and  r(A)=r(B)=a,
how can you deduct that x <=a^2, when a and x are arbitary?...

The claim is not correct. It requires A and B are commutative. However, many textbooks seems mistakely ignored this assumption. See wikipedia
http://en.wikipedia.org/wiki/Talk%3ASpectral_radius
and
https://chiasme.wordpress.com/20 ... ral-radius-formula/

But why they require A and B are commutative, I am not clear.
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22楼2014-12-15 10:49:15
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