Let V be a vector space of dimension n > 0 with a positive definite scalar product. Let A: V --> V be a linear map, symmetric with respect to the scalar product, and suppose that A has only one eigenvalue.
Now, we know that there exists an orthonormal basis of V consisting of eigenvectors: ${e_1, … e_n}$. Then we can represent our map A by a diagonal matrix with the eigenvalues along the diagonal. Note that each eigenvalue must be the same since A has only a single eigenvalue: $\lambda$
\begin{pmatrix} \lambda & 0 & \cdots & 0\\ \vdots & \ddots\\ 0 & 0 & \cdots & \lambda\\ \end{pmatrix}
Does this mean that every vector v in V is an eigenvector? Because if we represent the vector in the orthonormal basis, we immediately see that A(v) = $\lambda$v