• Support PF! Buy your school textbooks, materials and every day products Here!

Invertibility proof

  • Thread starter JG89
  • Start date
  • #1
726
1

Homework Statement


Let T be a linear operator on [tex] R^6 [/tex] whose characteristic polynomial is [tex] f(t) = (t-4)(t+1)^3 (t-2)^2[/tex]. Show that [tex] T^3 + 2T^2 -3T [/tex] is invertible.


Homework Equations





The Attempt at a Solution



Okay, T is invertible since 0 is not an eigenvalue of T. [tex] T^3 +2T^2 -3T [/tex] is invertible if and only if zero is not an eigenvalue of T, and this means that the only vector in the null space is the zero vector. Suppose that there was a non-zero vector, [tex] x' [/tex], in the null space. Then we have [tex] T^3(x') +2T^2(x') -3T(x') = 0 [/tex] then after applying [tex] T^{-1} [/tex] to both sides, we have [tex] T^3(x') + 2T^2(x') - 3T(x') = T^2(x') + 2T(x') -3x' \Rightarrow U(x') = T^2(x') + 2T(x') = 3x' [/tex] and so we see that 3 is an eigenvalue of U.

Now, I know from the characteristic polynomial that T's eigenvalues are [tex] \lambda_1 = 4, \lambda_2 = -1, \lambda_3 = 2 [/tex].

For [tex] \lambda_1 = 4 [/tex] corresponding to some vector x'_1,

[tex]U(x'_1) = T^2(x'_1) + 2T(x'_1) = T(4x'_1) + 8x'_1 = 16x'_1 + 8x'_1 = 24x'_1 [/tex].

Through similar methods, we can show that 24, -1, and 8 are eigenvalues of U.

Now here is where I get stuck. If this means that the ONLY eigenvalues of U is 24, -1, and 8, then since 3 isn't an eigenvalue of U, then T^3 + 2T^2 - 2T has no nonzero vector in its null-space and so it is invertible.

But if this means that 24, -1, and 8 are might only be some of the eigenvalues of U, then I can't conclude that 3 isn't an eigenvalue of U. I was thinking if I could prove that the algebraic multiplicity's of 24, -1, and 8 add up to 6, then since this is an operator on R^6 these would be the only eigenvalues of U, concluding my proof.

Any ideas?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
If you have (T^3+2T^2-3T)x=0, doesn't that mean (T+3)(T-1)Tx=0? Doesn't that mean T HAS to have a nontrivial eigenvector with eigenvalue either 0, 1 or -3? Aren't all of those incompatible with the characteristic polynomial? Aren't you overcomplicating this?
 
  • #3
726
1
I never even thought about factoring that. Wow
 

Related Threads for: Invertibility proof

  • Last Post
Replies
3
Views
15K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
11
Views
6K
Replies
1
Views
653
  • Last Post
Replies
8
Views
1K
Replies
5
Views
4K
  • Last Post
Replies
11
Views
3K
Top