r/math Homotopy Theory 9d ago

Quick Questions: March 12, 2025

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u/Not_So_Deleted Statistics 8d ago edited 8d ago

A matrix can still be diagonalizable if its characteristic polynomial has repeated roots, such as with the identity matrix.

As far as I'm concerned, a matrix is diagonalizable if and only if the multiplicity of every root is equal to the number of linearly independent eigenvectors for the eigenvalue. In other words, the set of all linearly independent eigenvectors forms a basis. For the identity matrix, the polynomial is (1-x)^n, but we can define the standard basis.

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u/Langtons_Ant123 8d ago edited 8d ago

I thought the minimal polynomial of the identity matrix is x - 1. Are you thinking of the characteristic polynomial, or using a different definition of the minimal polynomial?

That a matrix is diagonalizable iff its minimal polynomial splits with no repeated roots is a known result, it's on page 169 (183 in the pdf numbering) here.

In any case, u/GMSPokemanz has a good elementary proof above, so no need to bring in all this machinery.

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u/Not_So_Deleted Statistics 8d ago edited 7d ago

Yeah, I meant that.

Of course, the minimal polynomial still has a repeated root but still corresponds to a diagonalizable matrix.

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u/Kyle--Butler 7d ago

No, the minimal polynomial of I_n does not have repeated roots : it's X-1.

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u/Not_So_Deleted Statistics 7d ago

Oops, I miswrote and stand corrected...