For n > 1, the maximum multiplicity of any eigenvalue of an n × n matrix with real entries is?

The correct option/answer is n.

To understand why the maximum multiplicity of any eigenvalue of an $n\times n$ matrix is $n$, let us analyze the characteristic polynomial of a square matrix.
Let $A$ be an $n\times n$ matrix with real entries. The characteristic polynomial of $A$, denoted by $p\left(\lambda \right)$, is defined as:

$p\left(\lambda \right)=det(A-\lambda I)$

where $I$ is the $n\times n$ identity matrix, and $\lambda$ represents the eigenvalue variables. The determinant of an $n\times n$ matrix containing $\lambda$ along its diagonal yields a polynomial in $\lambda$ of degree exactly $n$.

By the Fundamental Theorem of Algebra, a polynomial of degree $n$ has exactly $n$ roots (counting multiplicity) in the complex number field. The roots of the characteristic polynomial are precisely the eigenvalues of the matrix.
Since the degree of the characteristic polynomial is $n$, the sum of the algebraic multiplicities of all eigenvalues of the matrix cannot exceed $n$.
Therefore, a single eigenvalue can have an algebraic multiplicity of at most $n$.

A simple example of a matrix achieving this maximum multiplicity is the identity matrix $I_n$ or any scalar matrix $c{I}_n$ where $c$ is a real number. For the matrix $c{I}_n$, the characteristic polynomial is:

$p\left(\lambda \right)={(c-\lambda )}^n$

Here, $\lambda =c$ is the only eigenvalue, and its multiplicity is exactly $n$.
Thus, the maximum multiplicity of any eigenvalue of an $n\times n$ matrix is indeed $n$.