Can Eigenvectors Diagonalize a Matrix with Only One Eigenvalue?

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The discussion centers on the diagonalization of a matrix with a single eigenvalue, specifically the matrix A defined as A = [[1, 2, 3], [0, 1, 2], [0, 0, 1]] over the fields R or C. The minimal polynomial of A is (x-1)^3, indicating that the only eigenvalue is 1. The primary decomposition theorem does not yield a decomposition in this case, as the number of linearly independent eigenvectors associated with the eigenvalue is insufficient for diagonalization.

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Let F= R or C, and A =
[1 2 3] is considered as linear operator in F3
[0 1 2]
[0 0 1]
then the minimal polynomial of A = (x-1)^3, can we say that the primary decomposition thm doesn't give any decomposition, can we find an invertible P s.t P^-1*A*p is a block diagonal matrix?
 
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it seems 1 is the only eigenvalue. now look for the eigenvectors with that eigenvalue. how many do you find? is that enough to diagonalize the matrix?
 

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