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Title: Algebraic Multiplicity with Fitting Index
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Series: Abstract Linear Algebra
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Chapter: Some matrix decompositions
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YouTube-Title: Abstract Linear Algebra 41 | Algebraic Multiplicity with Fitting Index
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Subtitle on GitHub: ala41_sub_eng.srt missing
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Timestamps (n/a)
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Subtitle in English (n/a)
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Quiz Content
Q1: Let $A \in \mathbb{C}^{n \times n}$ be a square matrix and $\lambda$ an eigenvalue with Fitting index $d$. We know that we have the block diagonalization for $N = A-\lambda \mathbf{1}$ with two invariant subspaces $U_1, U_2$. Let’s consider the linear map $\ell$ induced by $N$. What is correct?
A1: $(\ell|{U_1})^d = 0$ and $\ell|{U_2}$ is an isomorphism.
A2: $(\ell|{U_1})^{d-1} = 0$ and $\ell|{U_2}$ is surjective.
A3: $(\ell|{U_1})^{d-1}$ is an isomorphism and $\ell|{U_2}$ is injective.
A4: $(\ell|{U_1})^{d}$ is an isomorphism and $\ell|{U_2}$ is surjective.
Q2: Let $A \in \mathbb{C}^{n \times n}$ be a square matrix and $\lambda$ an eigenvalue with Fitting index $d$. What is always correct?
A1: The algebraic multiplicity of $\lambda$ is equal to $\mathrm{dim}(\mathrm{Ker}( A - \lambda \mathbf{1} )^d )$.
A2: The geometric multiplicity of $\lambda$ is equal to $\mathrm{dim}(\mathrm{Ker}( A - \lambda \mathbf{1} )^d )$.
A3: The algebraic multiplicity of $\lambda$ is equal to $\mathrm{dim}(\mathrm{Ker}( A - \lambda \mathbf{1} ) )$.
A4: The geometric multiplicities of all eigenvalues of $A$ are equal to $\mathrm{dim}(\mathrm{Ker}( A - \lambda \mathbf{1} ) )$.
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Date of video: 2025-02-16
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Last update: 2025-10
