I've just found a very strange exercise: from a 3x3 Matrix the eigenvalues are found to be 2: 1 with multiplicity 1 and the other with multiplicity 2; trying to find the kernel of the matrix A-LAMBDA I for the eigenvalue with multiplicity 2, it has dimension 1; the solution of the exercise otherwise, after have found the eigenvector from this matrix, calculates the square of A-LAMBDA I and then finds the eigenvector from this matrix and at the end uses this vector and the previous eigenvector to build a base for this eigenspace. What does this mean? How can this work?
2026-03-28 05:57:38.1774677458
Diagonalize a not diagonalizable matrix??
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