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Suppose $\bf A$ and $\bf B$ are $n \times m$ matrices with $n\ge m$, suppose $\bf AA^\top+BB^\top$ is invertible, and consider the matrix

$$\bf C = A^\top(AA^\top+BB^\top)^{-1}A.$$


Conjecture:

The matrix $\bf C$ has at least $n-m$ eigenvalues equal to $1$ and all other eigenvalues between $0$ and $1$


I am having a hard time proving this, so any help would be much appreciated. I have verified it numerically using the simple Matlab code bellow:

n = 7;
m = 5;
A = randn(n,m);
B = randn(n,m);
C = A'*((A*A'+B*B')^(-1))*A;
disp(eig(C))
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  • $\begingroup$ A minor modification is needed to avoid easy counterexamples. You should add the requirement $\text{rank}(A)=m$, and $\text{rank}(b)=m$, else the multiplicity of the eigenvalue $1$ can exceed $n-m$. With that fix, limited testing seems to support your conjecture, but I don't see how to prove it. Perhaps you meant "at least $n-m$ eigenvalues equal to $1$"? If so, then it's OK, but I think you should make clear that you mean at least $n-m$, rather than exactly $n-m$. $\endgroup$ – quasi Feb 28 '18 at 3:26
  • $\begingroup$ Also, can you give provide some context for the problem? Is it from a book? If not, how did the problem arise? $\endgroup$ – quasi Feb 28 '18 at 3:32
  • $\begingroup$ @quasi Thank you for your comment. I adjusted the conjecture. The problem is not from a book. Proving it would allow me to prove a property of the solution to a forecasting problem. $\endgroup$ – mzp Feb 28 '18 at 21:30
  • $\begingroup$ @quasi This is still perhaps not enough detail, but I think if I can prove this, I would be able to prove that ${\bf M}^n$ converges in this question: math.stackexchange.com/q/2660252/287326 $\endgroup$ – mzp Mar 1 '18 at 13:11
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$rank(AA^T)\leq m,rank(BB^T)\leq m$ implies that the symmetric $\geq 0$ matrices $AA^T,BB^T$ have at least $n-m$ zero eigenvalues, the other eigenvalues being $\geq 0$. Note also that $(AA^T+BB^T)^{-1}$ is symmetric $>0$ and, consequently, (1) the symmetric matrix $C$ is $\geq 0$.

On the other hand, (2) $spectrum(AA^T(AA^T+BB^T)^{-1})=spectrum(C)\cup \{0_1,\cdots,0_{n-m}\}$.

Since (3) $AA^T(AA^T+BB^T)^{-1}=I_n-BB^T(AA^T+BB^T)^{-1}$ and $rank(BB^T(AA^T+BB^T)^{-1})\leq m$, we deduce that $AA^T(AA^T+BB^T)^{-1}$ has at least $n-m$ eigenvalues equal to $1$ and, according to (2), $C$ too.

Since $BB^T(AA^T+BB^T)^{-1}$ is the product of a symmetrix matrix $\geq 0$ and a symmetric matrix $>0$, it has only $\geq 0$ eigenvalues and all the eigenvalues of $AA^T(AA^T+BB^T)^{-1}$ are $\leq 1$.

According to (1),(2), $spectrum(C)\subset [0,1]$.

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  • $\begingroup$ Thank you very much! Would you mind explaining how you obtained fact $(2)$? $\endgroup$ – mzp Mar 2 '18 at 14:29
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    $\begingroup$ (2) is a general result. Let $A\in M_{n,m},B\in M_{m,n},n\geq m$. Then $spectrum(AB)=spectrum(BA)\cup \{0_1,\cdots,0_{n-m}\}$. The symbol $\cup$ is used in the sense of a list. $\endgroup$ – loup blanc Mar 2 '18 at 15:45

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