For $A\in\mathbb{R}^{n\times n}$ and $B\in\mathbb{R}^{n\times k}$, does $(I_{n}-aA)^{-1}Bb$ determine $a\in\mathbb{R}$ and $b\in\mathbb{R}^k$?

Question

Let $A$ be a real $n\times n$ matrix, and $B$ be a real $n\times k$ matrix of rank $k<n$, with $\mathrm{col}(B)$ (the column space of $B$) not an invariant subspace of $A$. We assume $A\neq0,I_n$. Consider $$f(a,b)=(I_{n}-aA)^{-1}Bb,$$ for real scalar $a\in S$ and $b\in\mathbb{R}^k$. $S$ is the set of values of $a$ such that $(I_{n}-aA)$ is invertible.

We say that $f(a,b)$ identifies $a$ and $b$ over $\Omega$ if $f(a,b)=f(a^{\ast },b^{\ast})$ implies $(a,b)=(a^{\ast},b^{\ast})$ for any two pairs $(a,b),(a^{\ast},b^{\ast})\in\Omega$.

The problem is to describe the set over which $f(a,b)$ does not identify $a$ and $b$.

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What I have done so far:

Rewrite $f(a,b)=f(a^{\ast},b^{\ast})$ as $$ B(b-b^{\ast})+AB(ab^{\ast}-a^{\ast}b)=0.\tag{1}\label{1} $$ Thus, $f(a,b)$ identifies $a$ and $b$ over $\Omega$ if $\eqref{1}$ implies $(a,b)=(a^{\ast},b^{\ast})$ for all $(a,b),(a^{\ast},b^{\ast})\in\Omega$.

So far I understand only two particular cases:

1. If $\mathrm{rank}(B,AB)=2k$, then $\eqref{1}$ is satisfied iff $b-b^{\ast}=0$ and $ab^{\ast}-a^{\ast}b=0$, from which $(a,b)=(a^{\ast},b^{\ast})$ provided that $b\neq0$. Hence, in this case $f(a,b)$ identifies $a$ and $b$ over $S\times\mathbb{R}^{k}/\{0\}$.
2. Partition $B$ as $(B_{1},B_{2})$ where $B_{1}$ is $n\times k_{1}$ and $B_{2}$ is $n\times k_{2}$, with $0<k_{1}<k$, and suppose that $AB_{1}=B_{1}L$ for some $k_{1}\times k_{1}$ matrix $L$ ($\mathrm{col}(B_{1})$ is an invariant subspace of $A$), and that $\mathrm{rank}(B,AB)=k+k_{2}.$ Then, letting $\left( b_{1}^{\prime},b_{2}^{\prime}\right) $ be the partition of $b^{\prime}$ conformable with that of $B,$ $\eqref{1}$ becomes $$ B_{1}(b_{1}-b_{1}^{\ast}+L(ab_{1}^{\ast}-a^{\ast}b_{1}))+B_{2}(b_{2}% -b_{2}^{\ast})+AB_{2}(ab_{2}^{\ast}-a^{\ast}b_{2})=0. $$ Since the columns of $(B_{1},B_{2},AB_{2})$ are linearly independent, $\eqref{1}$ is satisfied if and only if $b_{1}-b_{1}^{\ast}+L(ab_{1}^{\ast }-a^{\ast}b_{1})=0$, $b_{2}-b_{2}^{\ast}=0$, and $ab_{2}^{\ast}-a^{\ast}% b_{2}=0$. Combining the second and third equalities, gives $b_{2}=b_{2}^{\ast }$ and $a=a^{\ast}$, provided that $b_{2}\neq0.$ Hence the first equality becomes $\left( I_{k_{1}}-aL\right) \left( b_{1}-b_{1}^{\ast}\right) =0$, which is equivalent to $b_{1}-b_{1}^{\ast}$ because $I_{k_{1}}-aL$ is invertible for any $a\in S$. This means that $f(a,b)$ identifies $(a,b)$ provided that not all entries of $b$ associated to columns of $B$ not in $\mathrm{col}(A)$ are zero (if they were we would essentially be in the case when $\mathrm{col}(B)$ is an invariant subspace of $A$--in that case it is clear that $\left( a,b\right) $ cannot be identified from $f(a,b)$)

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Update:

Let's look at the case $k=2.$ Write $B=(B_{1},B_{2})$, so $B_1$ and $B_2$ are vectors. Assume $\mathrm{rank}(B,AB)=\mathrm{rank}(B,AB_{2})=k+1$ (the case $\mathrm{rank}(B,AB)=2k$ is trivial, see case 1 above, and so is the case $\mathrm{rank}(B,AB)=k$), and write $$AB_{1}% =l_{1}B_{1}+l_{2}B_{2}+l_{3}AB_{2},$$ for $l_{1},l_{2},l_{3}\in\mathbb{R}$. From $\eqref{1}$, \begin{equation} \left[ \beta_{1}-\beta_{1}^{\ast}+l_{1}(a\beta_{1}^{\ast}-a^{\ast}\beta _{1})\right] B_{1}+\left[ \beta_{2}-\beta_{2}^{\ast}+l_{2}(a\beta_{1}^{\ast }-a^{\ast}\beta_{1})\right] B_{2}+\left[ a\beta_{2}^{\ast}-a^{\ast}\beta _{2}+l_{3}(a\beta_{1}^{\ast}-a^{\ast}\beta_{1})\right] AB_{2}=0. \end{equation} Since $\mathrm{rank}(B,AB_{2})=k+1,$ this equality is satisfied iff $$\left\{ \begin{array} [c]{c}% b_{1}-b_{1}^{\ast}+l_{1}(ab_{1}^{\ast}-a^{\ast}b_{1})=0\\ b_{2}-b_{2}^{\ast}+l_{2}(ab_{1}^{\ast}-a^{\ast}b_{1})=0\\ ab_{2}^{\ast}-a^{\ast}b_{2}+l_{3}(ab_{1}^{\ast}-a^{\ast}b_{1})=0 \end{array} \right.$$ As a linear system in the unknowns $b_{1}^{\ast},b_{2}^{\ast},a^{\ast}$, this is $$C\left[ \begin{array} [c]{c}% b_{1}^{\ast}\\ b_{2}^{\ast}\\ a^{\ast}% \end{array} \right] =\left[ \begin{array} [c]{c}% b_{1}\\ b_{2}\\ 0 \end{array} \right],\tag{1}\label{2}$$ where $$C=\left[ \begin{array} [c]{ccc}% 1-l_{1}a & 0 & l_{1}b_{1}\\ -l_{2}a & 1 & l_{2}b_{1}\\ -l_{3}a & a & l_{3}b_{1}+b_{2}% \end{array} \right] .$$

If $\mathrm{rank}(C)=3$, the only solution to the system is $\left( b^{\ast },a^{\ast}\right) =\left( b,a\right) $. If $\mathrm{rank}(C)<3$, there may be other solutions.

Note that $$ \det(C)=\left( l_{1}a-1\right) b_{2}-\left( l_{2}a+l_{3}\right) b_{1}. $$

Note that for fixed $a$, the set of $(b_1,b_2)$ such that $\det(C)=0$ is a line in $\mathbb{R}^2$.

Incidentally note that case 2. above (with $k=2$) obtains when $l_{2}=l_{3}=0.$ In that case, $\det(C)=\left( l_{1}a-1\right) b_{2}$, so $\det(C)=0$ only if $b_{2}=0$ ($l_{1}a-1=0$ is impossible as $l_{1}$ must be an eigenvalue of $A$, so $l_{1}a-1=0$ would contradict invertibility of $(I_n-aA)$). That is, in that case, we find again the result that $b,a$ is identifiable provided that $b_{2}\neq0$.

Answer 1

Here are some ideas.

We consider the relation $(I_n-aA)^{-1}Bb=v$ where $A,B$ and the vector $v$ are known and $a,b=[b_i]$ are unknown. Note that $n\geq k,\operatorname{rank}(B)=k$ and $1/a\notin \operatorname{spectrum}(A)$.

Then $Bb=(I-aA)v=v-aAv$, that is $Bb+aAv=v$. Let $B=[B_1,\cdots,B_k]$.

Assume that $A,B,v$ are randomly chosen. Then the $(B_i)_i,Av=w,v$ are random vectors in $\mathbb{R}^n$.

The above equation can be rewritten $\sum_{i\leq k}b_iB_i+aw=v$, that is, we want to decompose the vector $v$ on the $k+1$ vectors $(B_i)_i,w$; moreover, the decomposition must be unique.

It is possible with probability $1$ when $n=k+1$; otherwise, it is possible with probability $0$.

When $A$ is random, it's invertible with probability $1$. Here, $A$ is not assumed to be invertible and the question is: how to choose $v$ so that the decomposition exists and is unique? In particular, the vectors $v\in\ker(A)$ are not convenient. I think that there are many such subcases and I stand prudently on the sidelines.