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Let $X \in \mathbb{C}^{n \times m} $ be a rectangular matrix of full rank, and $X^*$ its hermitian conjugate, let $A \in \mathbb{C}^{m \times m}$ a square matrix, and let $f: \mathbb{C} \to \mathbb{C}$ be defined by $$ f(z) = \det \left( I_n + X \frac{1}{A - z I_m} X^* \right) $$ where $I_n$ is the $n \times n$ identity matrix.

Show that for $n \leq m$ solutions of $f(z)= 0$ are given by $z$ equal to an eigenvalue of $B : = A + X^* X$.

My attempt: for $n = m$, $f(z)$ can be straightforwardly rearranged to $$ f(z) = \det \left( X \frac{1}{B-A} \big( B - I_m z \big) \frac{1}{A - z I_m} X^* \right) = \frac{\det(X) \det(B - I_m z) \det(X^*)}{\det(A - z I_m) \det(B - A)} $$ where the desired result follows from the factor $\det(B - I_m z)$.

However I cannot factorise the determinant in this way for $n < m$, and I am unsure how to proceed.

ComptonScattering
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3 Answers3

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With the Sylvester determinant identity, we have $$ \begin{aligned} \det\left(I_n + \left( X\frac 1{A - zI_m}\right)X^*\right) &= \det\left(I_m + X^*\left( X\frac 1{A - zI_m}\right)\right) \\ & =\det\left(I_m + \big(X^*X \big) \frac 1{A - zI_m}\right) \\ & =\det\left(\big(A - zI_m + X^*X\big)\frac 1{A - zI_m}\right) \\ & =\frac{\det\Big(\big(A + X^*X\big) - zI_m)\Big)}{\det(A - zI_m)}. \end{aligned} $$

ComptonScattering
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Ben Grossmann
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By the Schur complement $$\det(z I_m-A)\,\det\left(I_n - X \frac{1}{z I_m-A} X^*\right) = \det(I)\det \left(z I_m-A-X^*X\right).$$ So $z$ should be the the eigenvalues of $A+X^*X$ but not of those of $A$.

Hans
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Lemma: $A$ is an $m\times n$ size matrix and $B$ is an $n\times m$ size matrix. $AB$ and $BA$ have the same nonzero eigenvalues.

Proof: Let $\lambda\neq0$ be an eigenvalue of $AB$ and $x$ a corresponding eigenvector, thus $ABx=\lambda x\neq0$. Obviously $Bx\neq0$. $$BA(Bx)=B(ABx)=B(\lambda x)=\lambda Bx.$$ So $\lambda$ is also an eigenvalue of $BA$ with an eigenvector $Bx$. In other words, the set of nonzero eigenvalues of $AB$ is contained in the set of nonzero eigenvalues of $BA$. The symmetric argument leads to the desired conclusion. $\quad\square$

I attempted to prove $AB$ and $BA$ have the same eigenvalues together with their algebraic multiplicities. After writing this above lemma, I realized that probably the best way to prove that is simply the Schur's complement. There is also another similar method to prove the same. In that case

But the above proposition is stronger than $\det(I+AB)=\det(I+BA)$. But then again we can just appeal to my other answer directly applying the Schur's complement.

Alternatively, we can confined ourselves to the narrow circumstance of this particular problem at hand, which only asks for the determinant to be zero. As @BenGrossman suggested, $\det(I+AB)=0 \iff AB$ has eigenvalue $-1 \iff BA$ has eigenvalue $-1\iff \det(I+BA)=0$. The rest proceeds via factorization by setting $A:=X$ and $B:=\frac 1{A - zI_m}X^*$.

Hans
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  • Actually that's a useful general result. Is there an enlightening proof? (other than running the arguments of your previous post backwards). – ComptonScattering Jan 05 '21 at 18:16
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    @ComptonScattering: Have you seen my addition of the lemma? – Hans Jan 05 '21 at 23:15
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    @ComptonScattering See also this post. That said, the proof in this answer is the one I prefer – Ben Grossmann Jan 06 '21 at 16:28
  • @Hans I don't see how we can go from $AB$ and $BA$ have the same non-zero eigenvalues to $I + AB$ and $I + BA$ have the same non-zero eigenvalues. I also don't see how we go from this statement to $\det(I + AB) = \det(I + BA)$. What I would have said is that, setting $P = X(A - zI)^{-1}$ and $Q = X$, we see that the first determinant is zero iff $-1$ is an eigenvalue of $PQ$, which occurs iff $-1$ is an eigenvalue of $QP$. This tells us that the determinants have the same zeros, which is what we wanted. Note that this doesn't tell us that the determinants are equal, at least not directly. – Ben Grossmann Jan 06 '21 at 16:32
  • @BenGrossmann: You are right. I was careless. This problem can indeed be based on the narrower argument you presented with regard to the sameness of the zeros of the determinants. However, $\det(I+AB)=\det(I+BA)$ is still correct, only that it needs an argument that is more elaborate. I will edit my answer later. – Hans Jan 07 '21 at 06:53