Show that each eigenvalue of matrix $M_Z \Sigma M_Z$ is less than the corresponding one of $M_X \Sigma M_X$ Let $Z:=[X \hspace{0.2cm} Y]$ be an $n\times k$ partitioned real matrix with rank $k$ and
$$M_X:=I_n-X(X'X)^{-1}X'$$
$$M_Z:=I_n-Z(Z'Z)^{-1}Z'$$
If $\Sigma$ is an $n\times n$ positive semi-definite matrix, how can I show that each eigenvalue of $M_Z \Sigma M_Z$ is less than the corresponding eigenvalue of $M_X \Sigma M_X$? (ordering the eigenvalues of each matrix in decreasing order)?
EDIT: I made the following progress:
Let $\Sigma^{1/2}$ denote the square root of $\Sigma$. If $A$ and $B$ are two square $n\times n$ matrices, then the characteristic polynomials of $AB$ and $BA$ coincide. Therefore the eigenvalues of $M_Z \Sigma M_Z$ and $M_X \Sigma M_X$ are identical to those of $\Sigma^{1/2} M_Z \Sigma^{1/2}$ and $\Sigma^{1/2} M_X \Sigma^{1/2}$ respectively (here I used the idempotence of $M_Z$ and $M_X$).
Now we have the following decomposition:
$$M_Z=M_X-P$$
(see here) where $P$ is a projection matrix. Plugging in we get
$$A=B+C$$
where $A:=\Sigma^{1/2} M_X \Sigma^{1/2},B:=\Sigma^{1/2} M_Z \Sigma^{1/2}$ and $C:=\Sigma^{1/2} P \Sigma^{1/2}$ are all positive semi-definite matrices.
From Weyl's inequalities we have
$$\lambda_i(A)\geq \lambda_i(B)+\lambda_n (C)\geq \lambda_i(B)  \hspace{1cm} i=1,\dots,n$$
which is what we wanted.
Is this ok?
 A: Note that $M_X$ and $M_Z$ are each orthogonal projections onto a subspaces of $\Bbb R^n$, and the image of $M_Z$ (the left nullspace of $Z$) is a subspace of the image of $M_X$ (the left nullspace of $X$).
Now, let $U$ denote a matrix whose columns form an orthonormal basis for the image of $M_Z$. Let $V$ be such that the columns of $[U \ \  V]$ form a basis of the image of $M_X$. We have
$$
M_Z = UU^T, \quad M_X = \pmatrix{U & V}\pmatrix{U&V}^T = UU^T + VV^T.
$$
Now, we note that whenever both products are defined, $AB$ and $BA$ have the same non-zero eigenvalues. Thus, the matrices
$$
M_Z(\Sigma M_Z), \quad \Sigma M_ZM_Z = \Sigma M_Z = (\Sigma U)U^T, \quad U^T \Sigma U=: P_Z
$$
will each have the same non-zero eigenvalues. Similarly, the matrices
$$
M_X \Sigma M_X, \quad \Sigma M_X = \Sigma \pmatrix{U & V} \pmatrix{U & V}^T, \\
\pmatrix{U & V}^T \Sigma \pmatrix{U & V} = \pmatrix{U^T\Sigma U & U^T \Sigma V\\ V^T \Sigma U & V^T \Sigma V} =: P_X
$$
will have the same non-zero eigenvalues.
However, note that $P_Z$ is a submatrix of $P_X$, which means that we can apply the Cauchy interlacing theorem. Thus, if we order the eigenvalues in decreasing order, we have
$$
\lambda_i(P_X) \geq \lambda_i(P_Z) \geq \lambda_{i + \ell}(X)
$$
where $\ell$ denotes the number of columns in $Y$.
