What are the criteria for determining if a matrix is diagonalizable?

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To determine if a matrix is diagonalizable, one must analyze its eigenvalues and their multiplicities. The algebraic multiplicity of the eigenvalue 0 in the characteristic polynomial λ(λ-5)(λ+2) is confirmed to be 1, similar to the other eigenvalues. For the polynomial (λ-2)(λ-4)(λ-α), the matrix A is diagonalizable if α does not equal 2 or 4, but if α equals either, it is necessary to check the geometric multiplicities of those eigenvalues. If there are two linearly independent eigenvectors for an eigenvalue with algebraic multiplicity greater than one, the matrix can still be diagonalizable. Understanding these concepts is crucial for determining diagonalizability in linear algebra.
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Homework Statement


1) Let's say I was trying to find the eigenvalues of a matrix and came up with the following characteristic polynomial:
λ(λ-5)(λ+2)
This would yield λ=0,5,-2 as eigenvalues. I'm kinda thrown off as to what the algebraic multiplicity of the eigenvalue 0 would be? I'm pretty sure it would just be 1 but I think I've misheard my instructor say otherwise during one example.

2) Let's say I have the following characteristic polynomial of matrix A.
(λ-2)(λ-4)(λ-α), where α is any number.

If I were trying to figure out values of α that make A diagonalizable, it would be any values of alpha that makes that particular eigenvalue different from 2 or 4. What if λ-α=0 did make λ equal 2 or 4? Do I have to check the geometric multiplicities for λ=2,4, or can I automatically assume that the matrix would not be diagonalizable.

Homework Equations


det(A-λI) = 0 where A is a matrix and λ are the eigenvalues A

The Attempt at a Solution

 
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Yes, the algebraic multiplicity of 0 would be 1. Just like 5 and -2. There's nothing special about 0. And for the second one if alpha=2 then you need to check if there are two linearly independent eigenvectors for the value 2. There might be and there might not be.
 
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