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When trying to find if a given matrix is diagonalizable, you need the check if for each Eigenvalue, the algebraic multiplicity equals the geometric multiplicity.

While the meaning of the geometric multiplicity is clear for me, ad the number of non dependent vectors the get multiply by the given eigenvalue, the meaning of the algebraic multiplicity is a bit unclear.

I know it's the number of appearances of the eigenvalue as a root in the characteristic polynomial.

So my question is, let's say we have a 2 transformations with 3 as an eigenvalue with the geometric multiplicity of 1. The first transformation an algebraic multiplicity of 2, while the second and algebraic multiplicity of 3.

what will be the difference in those to transformations? Thanks!

Omer811
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    One answer is that the second has a longer chain of generalized eigenvectors. Another answer is that, although $A - 3I$ has the same rank in both cases, $(A - 3I)^3$ does not. – Ben Grossmann Jul 10 '20 at 15:41
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    In any case, non-equality of algebraic and geometric multiplicities tells you that you have a non-diagonalizable transformation, or equivalently that you have some generalized eigenvectors. With that in mind, you might find my answer here to be useful. – Ben Grossmann Jul 10 '20 at 15:43

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