Solving for Eigenvalues and Determinant of a Singular Matrix

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The discussion revolves around solving for the eigenvalues and determinant of a singular 3x3 matrix. The characteristic polynomial is determined to be t(t-1)(t+3), indicating eigenvalues of 0, 1, and -3. It is concluded that A-3I is not invertible, as 3 is an eigenvalue, confirming the matrix's singularity. The determinant of the matrix is established as 0, while the trace is calculated to be -2 by summing the eigenvalues. The conversation also touches on the theorem that states a matrix with a zero eigenvalue is singular, reinforcing the findings.
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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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