Exponent of matrix/Diagonalization of matrix with repeated eigenvalue

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SUMMARY

The discussion focuses on the diagonalization and Jordan decomposition of a specific 3x3 matrix with repeated eigenvalues. The matrix in question has eigenvalues 1, 1, and 0, with corresponding eigenvectors (1, 0, 0), (0, 0, 0), and (0, 0, 1). It is established that the matrix cannot be diagonalized due to the lack of a complete set of independent eigenvectors, necessitating the use of Jordan Normal form instead. The user seeks to apply these concepts to find the matrix exponent of a related complex matrix.

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Hello,

it's been a while since i did linear algebra. i need some help. I have this matrix:

1 1 0
0 1 0
0 0 0.

I know the eigenvalues are 1,1,0; and that the eigenvectors will be: (1,0,0), (0,0,0) and (0,0,1). But I cannot do the jordan decomposition on the matrix i.e. write it in the form: P M P^-1. Where P is the matrix made up of eigenvectors, M is a diagonal matrix containing the eigenvalues: 1,1,0.

My main interest however, is in finding the matrix exponent of
1-2i 1+3i 0
0 1-2i 0
0 0 0.

If I can diagonalize the first matrix, I should be able to use the same method to diagonalize/jordan decompose this matrix so that i can find the matrix exponent.

Thanks for any help.
 
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Obviously, you cannot find P-1 because, taking the 0 vector as a column, you are forcing the determinant to be 0. In order to "diagonalize" a matrix you need to find a non-zero eigenvector (some texts do not accept the 0 vector as an eigenvector at all).

The difficulty is that not every matrix is diagonalizable! In order that a matrix be diagonalizable, there must exist a "complete set" of eigenvalues. That is, a basis for the space consisting of eigenvectors. For a three by three matrix, there must be three independent eigenvectors and that is not true here.

What you can do, if you cannot diagonalize a matrix, is put it in "Jordan Normal" form. Here, the matrix you are given is already in Jordan Normal form!
 

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