The Algebraic Multiplicity of Eigenvalues is Greater Than or Equal to Their Geometric Multiplicity
Theorem
A matrix matrix $A \in \mathbb{C}^{ m \times m}$ having an eigenvalue $\lambda$ with an algebraic multiplicity $a$ and a geometric multiplicity $g$ implies $a \ge g$.
Explanation
The algebraic and geometric multiplicities of an eigenvalue are not guaranteed to be the same. If they were, they wouldn’t have been defined differently in the first place. However, one certain thing is that regardless of how small the algebraic multiplicity may be, it is always greater than or equal to the geometric multiplicity.
Proof
Notation: For convenience, let’s refer to the row space of the given matrix’s basis as the matrix’s basis. When we say $m \in \mathbb{N}$, it means an identity matrix that has a size of $m \times m$. If the indices are omitted, it should be understood in the context of the operations involving matrices of corresponding sizes.
Given the assumption, the following holds:
$$ g = \dim \text{sp} \left\{ \mathbb{x}_{1} , \mathbb{x}_{2} , \cdots ,\mathbb{x}_{g} \right\} = \dim S_{\lambda} = \dim \left\{ \mathbb{x} \in \mathbb{C}^{ m } \ | \ A\mathbb{x} = \lambda \mathbb{x} \right\} $$
Meaning, with respect to $1 \le i \le g$, it is $A \mathbb{x}_{i} = \lambda \mathbb{x}_{i}$. Now, if the basis of the matrix $A$ is $$ \text{sp} \left\{ \mathbb{x}_{1} , \mathbb{x}_{2} , \cdots ,\mathbb{x}_{g} , \mathbb{y}_{1} , \mathbb{y}_{2} , \cdots , \mathbb{y}_{m-g} \right\} $$, then $P = \begin{bmatrix}\mathbb{x}_{1} & \mathbb{x}_{2} & \cdots & \mathbb{x}_{g} & \mathbb{y}_{1} & \mathbb{y}_{2} & \cdots & \mathbb{y}_{m-g} \end{bmatrix}$ is an invertible matrix.
$$ \begin{align*} AP =& A \begin{bmatrix} \mathbb{x}_{1} \cdots \mathbb{y}_{m-g} \end{bmatrix} \\ =& \begin{bmatrix} A \mathbb{x}_{1} & A \mathbb{x}_{2} & \cdots & A \mathbb{x}_{g} & A \mathbb{y}_{1} & A \mathbb{y}_{2} & \cdots & A \mathbb{y}_{m-g} \end{bmatrix} \\ =& \begin{bmatrix} \lambda \mathbb{x}_{1} & \lambda \mathbb{x}_{2} & \cdots & \lambda \mathbb{x}_{g} & A \mathbb{y}_{1} & A \mathbb{y}_{2} & \cdots & A \mathbb{y}_{m-g} \end{bmatrix} \\ =& \begin{bmatrix} \mathbb{x}_{1} \cdots \mathbb{y}_{m-g} \end{bmatrix} \begin{bmatrix} \lambda I_{g} & B \\ O & C \end{bmatrix} \end{align*} $$
Here, $B \in \mathbb{C}^{ g \times (m-g) }, C \in \mathbb{C}^{ (m-g) \times (m-g) }$ and $O$ is a $(m-g) \times g$ zero matrix. If we set $D = \begin{bmatrix} \lambda I_{g} & B \\ O & C \end{bmatrix}$, then $AP = PD$, and $P$ is an invertible matrix, therefore $$ A = PDP^{-1} $$ Also, since matrix $A$ and $D$ are similar, $$ \begin{align*} \det (A - \mu I) =& \det (D - \mu I) \\ =& \det \begin{bmatrix} (\lambda - \mu) I_{g} & B \\ O & C - \mu I_{m-g} \end{bmatrix} \\ =& (\lambda - \mu)^{g} \det ( C - \mu I_{m-g} ) \end{align*} $$
Among the roots fulfilling characteristic equation $\det (A - \mu I) = 0$, $\mu = \lambda$ has at least the geometric multiplicity of $g$. Hence, the algebraic multiplicity of $\lambda$, $a$, is greater than or equal to $g$.
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