Register to reply

Invertible linear transformation

by CarmineCortez
Tags: invertible, linear, transformation
Share this thread:
CarmineCortez
#1
Jan26-10, 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 || T-I || < 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(T-I) < 1.
Phys.Org News Partner Science news on Phys.org
Apple to unveil 'iWatch' on September 9
NASA deep-space rocket, SLS, to launch in 2018
Study examines 13,000-year-old nanodiamonds from multiple locations across three continents
JSuarez
#2
Jan26-10, 07:59 PM
P: 402
What would happen to T-I, if 0 was an eigenvalue of T? Is it compatible with the hypothesis?
CarmineCortez
#3
Jan26-10, 08:43 PM
P: 33
if 0 was an eigenvalue of T then T would be singular..

Dick
#4
Jan26-10, 08:56 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
Invertible linear transformation

Quote Quote by CarmineCortez View Post
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?
CarmineCortez
#5
Jan26-10, 09:12 PM
P: 33
Quote Quote by Dick View Post
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...
Dick
#6
Jan26-10, 09:19 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
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.
CarmineCortez
#7
Jan26-10, 10:04 PM
P: 33
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...
JSuarez
#8
Jan26-10, 10:18 PM
P: 402
For any norm [tex]\left\|v\right\|=\left\|-v\right\|[/tex]. Regarding the limit, remember the form of the geometric series.
JSuarez
#9
Jan26-10, 10:40 PM
P: 402
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.
CarmineCortez
#10
Jan28-10, 01:59 PM
P: 33
Quote Quote by JSuarez View Post
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...
JSuarez
#11
Jan29-10, 12:29 PM
P: 402
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?


Register to reply

Related Discussions
Is the linear transformation matrix T invertible Calculus & Beyond Homework 10
Matrices and Invertible Linear Transformations Precalculus Mathematics Homework 1
Linear Algebra - invertible matrix; determinants Calculus & Beyond Homework 2
Linear algebra-how do I know if something is invertible? Calculus & Beyond Homework 6
LINEAR ALG.: A^2 is invertible, is the inverse of a^2 equal to the square of inv(A)? Calculus & Beyond Homework 5