I am struggling with the following problem. In my current setting, I have two random matrices $X$ and $Y$ of dimensions $(T \times N)$ and a $(T \times 2)$ dimensional matrix $D$. The first column of $D$ is just a constant and the second contains a linear trend. However, the particular construction probably won't be important, just note that $T>2$. Then, I construct the idempotent matrix $M = I - D(D^\prime D)^{-1}D^\prime$, with $I$ being a $T$-dimensional identity matrix. Let $|| \cdot ||_2$ denote the spectral norm. Is it true that $||X^\prime M X - Y^\prime M Y||_2 \leq ||X^\prime X - Y^\prime Y||_2$ ? Simulations definitely indicate that the above inequality holds, no matter how I randomly generate $X,Y$. Of course, that does not count as a mathematical proof. If the above claim does not hold, is it then at least true that $||X^\prime M X - Y^\prime M Y||_2 \leq ||X^\prime X - Y^\prime Y||_F$, where $||\cdot||_F$ represents the Frobenius norm? My own reasoning only brings me to the following. It is straightforward to show that $||X^\prime M X||_2 = ||M XX^\prime||_2\leq ||M||_2||X^\prime X||_2 \leq ||X^\prime X||_2$, since $M$ is an idempotent matrix with the first $T-2$ eigenvalues equal to 1 and the remaining 2 equal to 0. The same obviously holds for $||Y^\prime M Y||$, but that does not seem sufficient for the claim to hold? For the Frobenius norm proof, I tried to write it as the trace of the inner product and use the inequality that $tr(AB) \leq ||A||_2\sum_i \sigma_i(B)$, but that got me nowhere. I really hope any of you can help. Please let me know if you need more information. Thanks! Etienne