Solving for Eigenvalues and Determinant of a Singular Matrix

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Homework Help Overview

The discussion revolves around a singular 3x3 matrix with specific properties related to its determinant and rank. The original poster attempts to find the characteristic polynomial, determine the invertibility of a matrix derived from the original, and calculate the determinant and trace of the matrix.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the eigenvalues of the matrix, with the original poster identifying 0, 1, and -3 as eigenvalues based on the given conditions. There is a question raised about the invertibility of A-3I and whether the reasoning provided is correct. Additionally, an alternative method for finding the trace is suggested, involving the sum of the eigenvalues.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's reasoning and exploring different interpretations of the results. Some participants question the implications of having a zero eigenvalue and its relation to the singularity of the matrix.

Contextual Notes

The original poster's approach is based on the properties of eigenvalues and the definitions of singular matrices. There is an underlying assumption regarding the relationship between eigenvalues and matrix invertibility that is being examined.

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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.
 
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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.
 

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