Solving for Eigenvalues and Determinant of a Singular Matrix

  • Thread starter Thread starter daniel_i_l
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
daniel_i_l
Gold Member
Messages
864
Reaction score
0

Homework Statement


You have a singular 3x3 matrix where det(A-I) = 0 , rank(A+3I) = 2
a) find the characteristic polynomial of A.
b) Is A-3I invertible?
c) Find det(A) and tr(A)


Homework Equations





The Attempt at a Solution


a) First I'll find the eigenvalues:
- 0 is obviously one since A is singular
-det(A-I)=0 => det(I-A) = 0 and so 1 is also an eigenvalue.
-rank(A+3I)=2 => rank(-3I - A) = 2 and so det(-3I - A) = 0 and so -3 is also one.
So the polynomial is t(t-1)(t+3)

b) No because if det(A-3I)=0 => det(3I-A)=0 => 3 is an eigenvalue but we already found 3 and that's the maximum.

c) det(A) =0 , and the polynomial is t(t^2+2t -3) = t^3 +2t^2 -3t and so tr(A) = 2.

Is that right? Am I missing anything?
Thanks.
 
Physics news on Phys.org
Basically fine, except that you argument in b) shows A-3I IS invertible, correct? And an alternative way to get the trace is just to sum the eigenvalues 0+1-3=-2. So since the characteristic polynomial is (t-e1)(t-e2)(t-e3) you can see that the coefficient of the t^2 is NEGATIVE the sum of the eigenvalues.
 
For part b:

Do you know the theorem that says that if you have any zero eigenvalue then the matrix is singular? If not, does it make sense to you why this is true?
 
Thanks everyone.