Let $$ A = \begin{bmatrix} 1&1&1&1\\ 1&1&0&0\\ 0&0&1&1\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix} $$ is $7 \times 4$ matrix of full column rank. For a positive definite matrix $\Sigma$, how can we prove that non zero eigenvalues of $A\Sigma A'$ and $A'A\Sigma$ are equal.
Note: $A'$ is the transpose of matrix $A$.
This is true for the nonzero eigenvalues:
Suppose first that $u \in \mathbb{R}^7$ is an eigenvector of $A\Sigma A'$ with eigenvalue $\lambda \neq 0$. Then $A'u \neq 0$ is an eigenvector of $A'A\Sigma$ with eigenvalue $\lambda$: $$ A'A\Sigma (A'u) = A'(A\Sigma A')u = A'(\lambda u) = \lambda (A'u). $$ On the other hand, suppose that $v \in \mathbb{R}^4$ is an eigenvector of $A'A\Sigma$ with eigenvalue $\mu$. Then $$ A\Sigma A'(A\Sigma v) = A\Sigma (A'A\Sigma v) = A\Sigma (\mu v) = \mu(A\Sigma v), $$ so provided that $A\Sigma v \neq 0$, it is an eigenvector of $A\Sigma A'$ with eigenvalue $\mu$. This follows because $\Sigma$ is positive-definite so has full rank, and $A$ has full rank, so each is injective onto its image.
However, $\ker{A'}$ is three-dimensional, so $A\Sigma A'$ also has $0$ as an eigenvalue, with three-dimensional eigenspace given by $\ker{A'}$.