 Solution

The premise contains a wording error: $\lambda$ must be a nonzero eigenvalue, rather than a nonzero eigenvector. Let $v\ne 0$ satisfyThen $Xv\ne 0$, since otherwise $X^{T}Xv=0$ would force $\lambda v=0$. Multiplying by $X$ givesThus $\boxed{\lambda \text{ is an eigenvalue of }X{X}^T}$, with eigenvector $Xv$. Conversely, applying $X^T$ maps an eigenvector of $X{X}^T$ with nonzero eigenvalue to one of $X^{T}X$.

These maps are inverses up to the factor $\lambda$ on the corresponding eigenspaces, so the nonzero eigenvalues have equal multiplicities. Their zero eigenvalues need not have the same multiplicity because the two matrices can have different sizes. For a unit eigenvector $v$ and real $X$, the normalized partner is $u=Xv/\sqrt{\lambda }$, since $\Vert Xv{\Vert }^2=v^T{X}^{T}Xv=\lambda >0$.