Diagonalisation of outer product matrices

For any $U\in\mathbb{R}^{n\times m}$ the matrix $A:=U\,U^\prime$ is symmetric with real eigenvalues. Write $U=(u_i^\mu)_{i=1\dots m}^{\mu=1\dots n}$. We are looking for eigenvectors $x$ to eigenvalue $\lambda$, i.e. $\forall_i\, u_i^\mu\, u_j^\mu\, x_j = \lambda\, x_i$. The ansatz $x=c^\nu\, u^\nu$ leads to

$$\forall_i\quad u_i^\prime\,\Delta\, c = \lambda\, u_i^\prime\, c$$ (90)

with $\Delta^{\mu\nu}=u^\mu\cdot u^\nu=(U^\prime\,U)^{\mu\nu}$. If $U$ has maximal rank, this is equivalent to $\Delta\, c = \lambda\, c$, i.e. it suffices to diagonalise $\Delta=U^\prime\, U$. The linear space $\left<u^\mu,\mu=1\dots n\right>^\bot$ is the eigenspace to eigenvalue $0$. The matrix $A:=U\,U^\prime$ has exactly $n$ (degeneracies counted) non-zero eigenvalues.