Invertible linear transformation

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Homework Help Overview

The problem involves proving the invertibility of a linear transformation T on R^n under the condition that the norm of T minus the identity matrix is less than 1. The discussion centers around properties of eigenvalues and spectral radius in relation to the invertibility of matrices.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of T being singular and the relationship between eigenvalues of T and T-I. There is discussion about the spectral radius and its connection to the convergence of series involving the transformation.

Discussion Status

The discussion is active with participants questioning assumptions about eigenvalues and singularity. Some have offered insights into the implications of the spectral radius being less than 1, while others are considering the convergence of series related to the transformation.

Contextual Notes

There are ongoing questions about the compatibility of the hypothesis with the existence of eigenvalues, as well as uncertainty regarding the properties of norms and their implications for convergence.

CarmineCortez
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Homework Statement


If T is a linear transformation on R^n with || T-I || < 1, prove that T is invertible.



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(T-I) < 1.
 
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What would happen to T-I, if 0 was an eigenvalue of T? Is it compatible with the hypothesis?
 
if 0 was an eigenvalue of T then T would be singular..
 
CarmineCortez said:
if 0 was an eigenvalue of T then T would be singular..

Ok, so if T is not invertible then Tv=0 for some v. So v corresponds to what eigenvalue of T-I?
 
Dick said:
Ok, so if T is not invertible then Tv=0 for some v. So v corresponds to what eigenvalue of T-I?

0 = λ*v + I*v
=> -1 = λ

but I know my spectral radius is <1 so contradiction...
 
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.
 
I need to show: sum from k=0 to infinity of (I-T)^k converges absolutely to T^(-1)

so if ||T-I|| <1 then is ||I-T|| < 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...
 
For any norm [tex]\left\|v\right\|=\left\|-v\right\|[/tex]. Regarding the limit, remember the form of the geometric series.
 
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
JSuarez said:
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.

There is a thm that says if spectral norm <1 then A^n -> 0 as n-> infinity.

and I proved above that spectral norm is <1

so I'm lost again...
 
Last edited:
  • #11
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 = I-T?
 

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