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suppose I have vectors $(x_i,y_i)_{i=1}^n$ where $x_i \in [0,1]^n$ and $y_i \in [0,1]^m$. I know that $Ax_i = y_i$

In this case, is there any way or any paper that I can recover matrix operator $A$?

I have checked out some papers discussing about low rank matrix recovery, but it seems I am dealing with a different question.

user1292919
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1 Answers1

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Let $X$ denote the matrix whose columns are $x_1, \dots, x_n$, and let $Y$ denote the matrix whose columns are $y_1,\dots,y_n$. You can rewrite your statement on the $x_i$ and $y_i$ as $$ A X = Y $$ In your case, the matrix $X$ has $n$ columns and $n$ rows. This equation will allow you to recover $A$ if and only if the columns $x_i$ are linearly independent; in this case we can compute $$ A = YX^{-1} $$ More generally, we can cover $A$ as long as the vectors $x_i$ span $\Bbb R^n$, even if there are more than $n$ of them.

Ben Grossmann
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  • Suppose if we have $k<n$ number of $x_i$'s. Can we recover some part of A or are we hopeless? – user1292919 May 09 '18 at 14:34
  • @user1292919 in that case, we can only recover what $A$ does over the column space of $X$. In particular: if $x$ is a vector in the column-space of $X$, then we can find a vector $u$ such that $Xu = b$. From there, we have $$ Ab = AXu = Yu $$ if $x$ is not in that column space, however, we are indeed "hopeless". – Ben Grossmann May 09 '18 at 15:49