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

The correct option is n.

To understand why the maximum multiplicity of any eigenvalue of an $n\times n$ matrix is $n$, let us analyze the characteristic equation of the matrix.

Let $A$ be an $n\times n$ square matrix with real entries. The eigenvalues of $A$ are the roots of its characteristic polynomial, which is defined by:
$$p\left(\lambda \right)=det\left(A-\lambda I\right)$$
where $I$ is the $n\times n$ identity matrix and $\lambda$ represents the eigenvalue.

Since $A$ is of size $n\times n$, the characteristic polynomial $p\left(\lambda \right)$ is a polynomial of degree exactly $n$ in the variable $\lambda$. According to the Fundamental Theorem of Algebra, a polynomial of degree $n$ has exactly $n$ roots (counting multiplicity) over the complex field.

If ${\lambda }_1$ is a root of the polynomial, we can factor the characteristic polynomial as:
$$p\left(\lambda \right)={\left(\lambda -{\lambda }_1\right)}^{k}q\left(\lambda \right)$$
where $k$ is the algebraic multiplicity of the eigenvalue ${\lambda }_1$, and $q\left(\lambda \right)$ is a polynomial of degree $n-k$.

Since the degree of the entire polynomial $p\left(\lambda \right)$ is $n$, the algebraic multiplicity $k$ of any single eigenvalue cannot exceed $n$ (meaning $k\le n$).

An example that achieves this maximum multiplicity is the identity matrix $I_n$. The characteristic polynomial for $I_n$ is:
$$p\left(\lambda \right)={\left(1-\lambda \right)}^n$$
Here, the single eigenvalue $\lambda =1$ has a multiplicity of exactly $n$. Therefore, the maximum multiplicity of any eigenvalue of an $n\times n$ matrix is $n$.