What is wrong with this diagonalization problem.

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The discussion revolves around a diagonalization problem involving a matrix A with eigenvalues 5, 3, and 3. The user initially struggles with the concept of diagonalization due to having linearly dependent eigenvalues, which complicates finding an invertible matrix P and a diagonal matrix D. It is clarified that the matrix is indeed diagonalizable if the minimal polynomial has distinct eigenvalues, and that generalized eigenvectors can be used when eigenvalues are repeated. The user ultimately resolves their confusion by recognizing that the eigenvalue 3 actually corresponds to two independent eigenvectors. The thread concludes with the user expressing gratitude for the assistance received.
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Homework Statement


A is

[4 0 1
2 3 2
1 0 4]

Find an invertible P and a diagonal D so that D=P-1AP.

I keep getting two linearly dependent eigenvalues which means it's not diagonal but this problem doesn't state "If it can't be done explain why" or anything like that. So I just want to verify with some of you.

The Attempt at a Solution


I subtract with LI and take the determinant and get:

(L-3)((L-4)^2 - 1)
(L-3)(L^2-8L+16)-(L-3)
L^3-8L^2+16L-3L^2+24L-48-L+3
L^3-11L^2+39L-45

Which I then factor out to be 5, 3, 3.

Am I doing something wrong/missing something?
 
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The eigenvalues are correct. Now, did you plug each eigenvalue, one at a time, into your matrix to find the corresponding eigenvectors? There are two eigenvectors for eigenvalue 3, and one for eigenvalue 5.
 
Right but you can't find the diagonalization when the two eigenvalues are the same. They are linearly dependent. Right? I just wanted to verify that I wasn't going crazy.

Once you have two identical eigenvectors for P you can't invert it to eventually find D. Or am I supposed to just assume that D is the diagonal matrix of 5,3,3 even though you can't actually get to that point.
 
The minimal polynomial of your matrix (x - 5)(x - 3), so it is diagonalizable.
 
It is diagonalizable if and only if the characteristic polynomial of your matrix, A, splits and for each eigenvalue X of A, the multiplicity of X equals n - rank (A-XI). So check to see if these criteria apply here (you have almost done the first step)
 
When two or more of your eigenvalues are the same, you need to search for generalized eigenvectors. These generalized eigenvectors will be linearly independent, and you can construct your P-matrix from them.
 
VeeEight said:
It is diagonalizable if and only if the characteristic polynomial of your matrix, A, splits and for each eigenvalue X of A, the multiplicity of X equals n - rank (A-XI). So check to see if these criteria apply here (you have almost done the first step)

Or more simply, if and only if the minimal polynomial has the form (x - a1)(x - a2)...(x - an) where ai are the distinct eigenvalues of A. D will be the diagonal matrix having a1, a2, etc. down the diagonal and P will be the matrix whose columns are a linearly independent set of eigenvectors corresponding to the eigenvalues of A.
 
Can one of you help me find the generalized vectors I have no idea what to do.
 
Nm, I figured it out. when looking for the eigenvector for 3 it actually splits into 2. I think I did the math wrong the first time. Thanks you guys.
 

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