**Proof plan.** The lower bound $m_\lambda \geq 1$ is immediate. For the upper bound $m_\lambda \leq M_\lambda$, we choose a basis adapted to the eigenspace $E_\lambda$ and compute the characteristic polynomial in that basis, showing that $(t - \lambda)^{m_\lambda}$ divides $\chi_A(t)$. **Step 1: Lower bound.** Since $\lambda$ is an eigenvalue, there exists $v \neq \mathbf{0}$ with $Av = \lambdav$. Then $v \in E_\lambda$, so $\dim E_\lambda \geq 1$, i.e., $m_\lambda \geq 1$. **Step 2: Upper bound.** Let $m = m_\lambda$ and choose a basis $\{u_1, \ldots, u_m\}$ for $E_\lambda$. Extend to a basis $\{u_1, \ldots, u_m, w_1, \ldots, w_{n-m}\}$ for the full space. In this basis, $Au_i = \lambdau_i$ for $i = 1, \ldots, m$, so the matrix of $A$ has the block form: \begin{align*} [A] = \begin{pmatrix} \lambda I_m & B \\ 0 & C \end{pmatrix}, \end{align*} where $I_m$ is the $m \times m$ identity, $B$ is $m \times (n-m)$, and $C$ is $(n-m) \times (n-m)$. The zero block in the lower-left arises because $Au_i = \lambdau_i$ has no component along the $w_j$ directions. **Step 3: Compute the characteristic polynomial.** Using the block triangular structure: \begin{align*} \chi_A(t) = \det([A] - tI) = \det\begin{pmatrix} (\lambda - t)I_m & B \\ 0 & C - tI_{n-m} \end{pmatrix} = (\lambda - t)^m \det(C - tI_{n-m}). \end{align*} The factor $(\lambda - t)^m$ shows that $\lambda$ is a root of $\chi_A(t)$ with multiplicity at least $m$. That is, $M_\lambda \geq m_\lambda$.