[proofplan] We establish the two inequalities separately. The lower bound $1 \le m_g(\lambda)$ follows from the definition of eigenvalue: the eigenspace is nontrivial. The upper bound $m_g(\lambda) \le m_a(\lambda)$ is proved by choosing a basis for the eigenspace, extending it to a basis of $\mathbb{F}^n$, and examining the characteristic polynomial of the resulting matrix representation. [/proofplan] [step:Establish the lower bound $1 \le m_g(\lambda)$] Since $\lambda$ is an eigenvalue of $A$, there exists a nonzero vector $v \in \mathbb{F}^n$ with $Av = \lambda v$, so $v \in \ker(A - \lambda I)$. Therefore $\ker(A - \lambda I) \neq \{0\}$, which gives \begin{align*} m_g(\lambda) = \dim \ker(A - \lambda I) \ge 1. \end{align*} [guided] What does it mean for $\lambda$ to be an eigenvalue? By the [Eigenvalues as Roots of the Characteristic Polynomial](/theorems/918), $\lambda$ is a root of the characteristic polynomial $p_A$, and equivalently there exists a nonzero vector $v \in \mathbb{F}^n$ satisfying $Av = \lambda v$. Rewriting this as $(A - \lambda I)v = 0$, we see that $v$ lies in the kernel $\ker(A - \lambda I)$. Define the eigenspace $E_\lambda := \ker(A - \lambda I)$. Since $v \neq 0$ and $v \in E_\lambda$, the eigenspace contains at least one nonzero vector, so $E_\lambda \neq \{0\}$. The geometric multiplicity is defined as $m_g(\lambda) = \dim E_\lambda$. Since $E_\lambda$ is a nonzero subspace of $\mathbb{F}^n$, its dimension is at least $1$. Therefore \begin{align*} m_g(\lambda) = \dim E_\lambda \ge 1. \end{align*} This lower bound is sharp: it is attained whenever the eigenspace is one-dimensional, for instance when $A$ has $n$ distinct eigenvalues. [/guided] [/step] [step:Choose a basis for the eigenspace and extend it to $\mathbb{F}^n$] Let $m := m_g(\lambda)$ and let $\{v_1, \ldots, v_m\}$ be a basis for $\ker(A - \lambda I)$. By the [Extension to a Basis](/theorems/3264), extend this to a basis $\mathcal{B} = \{v_1, \ldots, v_m, v_{m+1}, \ldots, v_n\}$ of $\mathbb{F}^n$. Let $P \in \mathbb{F}^{n \times n}$ be the invertible matrix whose columns are $v_1, \ldots, v_n$. [guided] We want to see how the eigenspace structure constrains the characteristic polynomial. The strategy is to choose coordinates that make the eigenspace visible: if we can arrange for the first $m$ basis vectors to be eigenvectors, the matrix representation will have a revealing block structure. Let $m := m_g(\lambda)$ and let $\{v_1, \ldots, v_m\}$ be a basis for $E_\lambda = \ker(A - \lambda I)$. Since $E_\lambda$ is a subspace of $\mathbb{F}^n$, the [Dimension of Subspaces](/theorems/375) gives $m \le n$. These $m$ vectors are linearly independent by definition of a basis for $E_\lambda$. We now extend this to a full basis of $\mathbb{F}^n$. By the [Extension to a Basis](/theorems/3264), any linearly independent set in a finite-dimensional vector space can be extended to a basis. Applying this to $\{v_1, \ldots, v_m\}$, we obtain vectors $v_{m+1}, \ldots, v_n \in \mathbb{F}^n$ such that $\mathcal{B} = \{v_1, \ldots, v_m, v_{m+1}, \ldots, v_n\}$ is a basis of $\mathbb{F}^n$. Note that the extension vectors $v_{m+1}, \ldots, v_n$ are not eigenvectors in general — they are chosen only to complete the basis. Form the change-of-basis matrix $P = [v_1 \mid \cdots \mid v_n] \in \mathbb{F}^{n \times n}$, whose columns are the basis vectors. Since the columns of $P$ form a basis of $\mathbb{F}^n$, they are linearly independent, so $P$ is invertible by the [Determinant Invertibility Criterion](/theorems/396). The similar matrix $B := P^{-1}AP$ represents $A$ in the basis $\mathcal{B}$. [/guided] [/step] [step:Compute the matrix representation in the new basis] Since $Av_j = \lambda v_j$ for $j = 1, \ldots, m$, the matrix $B := P^{-1}AP$ has the block form \begin{align*} B = P^{-1}AP = \begin{pmatrix} \lambda I_m & C \\ 0 & D \end{pmatrix}, \end{align*} where $I_m$ is the $m \times m$ identity matrix, $C \in \mathbb{F}^{m \times (n-m)}$, and $D \in \mathbb{F}^{(n-m) \times (n-m)}$. The lower-left block is zero because the first $m$ columns of $B$ are $P^{-1}Av_j = P^{-1}(\lambda v_j) = \lambda e_j$ for $j = 1, \ldots, m$, where $e_j$ denotes the $j$-th standard basis vector in $\mathbb{F}^n$. [guided] Now we compute the representation of $A$ in the basis $\mathcal{B}$. The similar matrix is $B := P^{-1}AP$. For $j = 1, \ldots, m$, the $j$-th column of $B$ is \begin{align*} Be_j = P^{-1}Av_j = P^{-1}(\lambda v_j) = \lambda P^{-1} v_j = \lambda e_j, \end{align*} since $P^{-1}v_j = e_j$ (the $j$-th standard basis vector) by construction. So the first $m$ columns of $B$ are $\lambda e_1, \ldots, \lambda e_m$. This means $B$ has the block upper-triangular form \begin{align*} B = \begin{pmatrix} \lambda I_m & C \\ 0 & D \end{pmatrix}, \end{align*} where $C \in \mathbb{F}^{m \times (n-m)}$ records the coordinates of $Av_{m+1}, \ldots, Av_n$ in the $v_1, \ldots, v_m$ directions, and $D \in \mathbb{F}^{(n-m) \times (n-m)}$ records them in the $v_{m+1}, \ldots, v_n$ directions. The zero block in the lower-left arises precisely because each $Av_j$ ($j \le m$) is a scalar multiple of $v_j$ and contributes nothing to the $v_{m+1}, \ldots, v_n$ components. [/guided] [/step] [step:Factor the characteristic polynomial to extract $(\lambda - t)^m$] The [Characteristic Polynomial Is a Similarity Invariant](/theorems/402), so $p_A(t) = p_B(t)$. The block upper-triangular structure gives \begin{align*} p_A(t) = \det(B - tI) = \det\begin{pmatrix} (\lambda - t)I_m & C \\ 0 & D - tI_{n-m} \end{pmatrix} = \det\bigl((\lambda - t)I_m\bigr) \cdot \det(D - tI_{n-m}) = (\lambda - t)^m \cdot p_D(t), \end{align*} where the second equality uses the [Block Triangular Determinant](/theorems/399) formula. Since $p_D(t)$ is a polynomial in $t$, the factor $(\lambda - t)^m$ divides $p_A(t)$. By definition, $m_a(\lambda)$ is the largest power of $(\lambda - t)$ dividing $p_A(t)$, so \begin{align*} m_g(\lambda) = m \le m_a(\lambda). \end{align*} [guided] Since similar matrices have the same characteristic polynomial by the [Characteristic Polynomial Is a Similarity Invariant](/theorems/402), we have $p_A(t) = p_B(t)$. Now we exploit the block structure. Applying the [Block Triangular Determinant](/theorems/399) to $B - tI$: \begin{align*} p_B(t) = \det(B - tI) = \det\begin{pmatrix} (\lambda - t)I_m & C \\ 0 & D - tI_{n-m} \end{pmatrix} = \det\bigl((\lambda - t)I_m\bigr) \cdot \det(D - tI_{n-m}). \end{align*} The first factor is $(\lambda - t)^m$ and the second is $p_D(t)$, the characteristic polynomial of the $(n-m) \times (n-m)$ matrix $D$. Therefore \begin{align*} p_A(t) = (\lambda - t)^m \cdot p_D(t). \end{align*} The algebraic multiplicity $m_a(\lambda)$ is the multiplicity of $\lambda$ as a root of $p_A$. Since $(\lambda - t)^m$ divides $p_A(t)$, and $p_D(t)$ may contribute additional factors of $(\lambda - t)$, we conclude $m_a(\lambda) \ge m = m_g(\lambda)$. Combining with the lower bound from the first step gives $1 \le m_g(\lambda) \le m_a(\lambda)$. [/guided] [/step]