
#1
Jan2610, 07:03 PM

P: 33

1. The problem statement, all variables and given/known data
If T is a linear transformation on R^n with  TI  < 1, prove that T is invertible. 3. The attempt at a solution So a linear transformation T is invertible iff the matrix T is not singular. and I know for any matrix A, A > spectral radius(A). so, spectral radius(TI) < 1. 



#2
Jan2610, 07:59 PM

P: 403

What would happen to TI, if 0 was an eigenvalue of T? Is it compatible with the hypothesis?




#3
Jan2610, 08:43 PM

P: 33

if 0 was an eigenvalue of T then T would be singular..




#4
Jan2610, 08:56 PM

Sci Advisor
HW Helper
Thanks
P: 25,167

Invertible linear transformation 



#5
Jan2610, 09:12 PM

P: 33

=> 1 = λ but I know my spectral radius is <1 so contradiction... 



#6
Jan2610, 09:19 PM

Sci Advisor
HW Helper
Thanks
P: 25,167

Yes. That's it. You could also say (T+I)v=(v) means T+I>=1 and not even say anything about spectral radius. Still a contradiction with T+I<1.




#7
Jan2610, 10:04 PM

P: 33

I need to show: sum from k=0 to infinity of (IT)^k converges absolutely to T^(1)
so if TI <1 then is IT < 1? and all the properties I listed carry over? I'm still not too sure where to go with this. when the spectral radius is <1, the higher powers of the matrix tend to 0, so it clearly converges... 



#8
Jan2610, 10:18 PM

P: 403

For any norm [tex]\left\v\right\=\left\v\right\[/tex]. Regarding the limit, remember the form of the geometric series.




#9
Jan2610, 10:40 PM

P: 403

In fact, it's easier if you consider a matrix [tex]S[/tex], with [tex]\left\S\right\<1[/tex] and prove that:
[tex]\sum_{n=0}^{\infty}S^n[/tex] Converges absolutely and compute the limit. 



#10
Jan2810, 01:59 PM

P: 33

and I proved above that spectral norm is <1 so I'm lost again... 



#11
Jan2910, 12:29 PM

P: 403

What can you say about the real series:
[tex] \sum_{n=0}^{\infty}\left\S\right\^n [/tex] When [tex]\left\S\right\<1[/tex]? Does it converge? if yes, what's the sum? Is it related to ypur original series if S = IT? 


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