How could I prove that if the geometric multiplicity of all eigen values of a matrix equals its algebraic multiplicity, then the matrix is diagonalizable?
2026-03-24 20:46:31.1774385191
Diagonalization:Algebraic and Geometric multiplicity
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Well, that's obvious: if the geometric multiplicities of the eigenvalues are equal to their algebraic multiplicities, as the sum of the algebraic multiplicities is the dimension $n$ of the vector space, it means we have a set of $n$ linearly independent eigenvectors, which are therefore a basis of eigenvectors.