why geometric multiplicity is bounded by algebraic multiplicity?

The algebraic multiplicity of λi is the degree of the root λi in the polynomial det(Aλ).
The geometric multiplicity is the dimension of the eigenspace of eigenvalue λi.

For example:
[1101] has root 1 with algebraic multiplicity 2, but the geometric multiplicity 1.

My question is : why geometric multiplicity is always bounded by algebraic multiplicity?

Thanks.

Answer

Suppose the geometric multiplicity of the eigenvalue λ of A is k. Then we have k linearly independent vectors v1,,vk such that Avi=λvi. If we change our basis so that the first k elements of the basis are v1,,vk, then with respect to this basis we have
A=(λIkB0C)
where Ik is the k×k identity matrix. Since the characteristic polynomial is independent of choice of basis, we have
charA(x)=charλIk(x)charC(x)=(xλ)kcharC(x)
so the algebraic multiplicity of λ is at least k.

Attribution
Source : Link , Question Author : Jack2019 , Answer Author : Alex Becker

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