Orthogonal matrix multiplied by diagonal matrix multiplied by transpose of the orthogonal matrix

Question

Suppose tall matrix $A$ is $n \times k$ and that its columns are orthogonal, i.e., $A' A = I_k$. Suppose further that diagonal $M$ is $n \times n$ and has either $1$ or $0$ on its main diagonal. Suppose also that $n \gg k$ and $\mbox{rank} (M) > k$. Is it true that $A'MA = I_k$? If so, why?

I have gotten as far as:

$$A'MA = A'M'MA = (MA)'(MA)$$

but am unsure if $M A$ should also have orthogonal columns and, if so, why. If this is not true, are there any similar statements I could make about $A' M A$?

Answer 1

No, consider $n=k$ then the rank of $A’MA$ equals the rank of $M$, which only equals $k$ if $M$ is the identity.

Edit: Still no. This question is the same as the following:

I have $k$ many $n$-dimensional orthonormal vectors, which I truncate down to their first $m$ dimensions - are the resulting $m$-dimensional vectors still orthonormal?

Of course the answer to this is no. Take $k$ many $m$-dimensional orthogonal (not orthonormal) vectors (possible since $m>k$) but which have norms $<1$, then fill up the remaining $n-m$ entries in any way you please such that the results are still orthogonal, but the vectors have norm $=1$. As long as $n-m \geq k$, there is enough room to do this. You get a matrix $A$ such that $0 < A'MA < I_k$.

Note this is just one possible counterexample, it is also possible that the truncation to $m$-dimensions is not even still orthogonal! For instance the truncations could even be zero.

Or consider how $(1,1,1,1), (1,1,-1,-1)$ are orthogonal, but the truncated vectors $(1,1)$ and $(1,1)$ are not.

Answer 2

If I understood you correctly, the diagonal entries of $M$ is either $0$ or $1$. If you let $M=diag(0\cdots 0)$ then $A'MA=$ zero matrix.