How to undo a matrix-vector multiplication












0












$begingroup$


I have an iterative algorithm that computes a matrix-vector multiplication such as:



$$
b = Av
$$



I know the vector b (which is the result of the algorithm) and the vector v. Is there a way to get the matrix A?






matrices vectors inner-product-space







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you know just one pair of vectors $mathbb b$ and $mathbb v,$ or more than one pair? Do you have the ability to provide several examples of $mathbb v$ and get the resulting vector $mathbb b$ in each case?
    $endgroup$
    – David K
    Dec 26 '18 at 22:09
















0












$begingroup$


I have an iterative algorithm that computes a matrix-vector multiplication such as:



$$
b = Av
$$



I know the vector b (which is the result of the algorithm) and the vector v. Is there a way to get the matrix A?






matrices vectors inner-product-space







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you know just one pair of vectors $mathbb b$ and $mathbb v,$ or more than one pair? Do you have the ability to provide several examples of $mathbb v$ and get the resulting vector $mathbb b$ in each case?
    $endgroup$
    – David K
    Dec 26 '18 at 22:09














0












0








0


1



$begingroup$


I have an iterative algorithm that computes a matrix-vector multiplication such as:



$$
b = Av
$$



I know the vector b (which is the result of the algorithm) and the vector v. Is there a way to get the matrix A?






matrices vectors inner-product-space







share|cite|improve this question









$endgroup$




I have an iterative algorithm that computes a matrix-vector multiplication such as:



$$
b = Av
$$



I know the vector b (which is the result of the algorithm) and the vector v. Is there a way to get the matrix A?






matrices vectors inner-product-space




matrices vectors inner-product-space






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 26 '18 at 11:20









Kart AlexiusKart Alexius

407




407












  • $begingroup$
    Do you know just one pair of vectors $mathbb b$ and $mathbb v,$ or more than one pair? Do you have the ability to provide several examples of $mathbb v$ and get the resulting vector $mathbb b$ in each case?
    $endgroup$
    – David K
    Dec 26 '18 at 22:09


















  • $begingroup$
    Do you know just one pair of vectors $mathbb b$ and $mathbb v,$ or more than one pair? Do you have the ability to provide several examples of $mathbb v$ and get the resulting vector $mathbb b$ in each case?
    $endgroup$
    – David K
    Dec 26 '18 at 22:09
















$begingroup$
Do you know just one pair of vectors $mathbb b$ and $mathbb v,$ or more than one pair? Do you have the ability to provide several examples of $mathbb v$ and get the resulting vector $mathbb b$ in each case?
$endgroup$
– David K
Dec 26 '18 at 22:09




$begingroup$
Do you know just one pair of vectors $mathbb b$ and $mathbb v,$ or more than one pair? Do you have the ability to provide several examples of $mathbb v$ and get the resulting vector $mathbb b$ in each case?
$endgroup$
– David K
Dec 26 '18 at 22:09










2 Answers
2






active

oldest

votes


















1












$begingroup$

The first column of $A$ is $A.(1,0,0,ldots,0)$, the second column is $A.(0,1,0,ldots,0)$ and so on.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well the algorithm starts with $A_0 = gamma I$ so your answer is kind of correct. But I wanted to know $A$ after the algorithm, given that I know the result $Av$ and $v$. If it helps I am referring to the algorithm called two-loop recursion of Quasi-Newton method.
    $endgroup$
    – Kart Alexius
    Jan 6 at 15:05










  • $begingroup$
    I don't know. My suggestion is that you post a new question, but this time don't forget to mention that algorithm.
    $endgroup$
    – José Carlos Santos
    Jan 6 at 15:09





















1












$begingroup$

There exists an infinity of solutions to the problem
$$b=Av$$
Where $b$ and $v$ are vectors



The problem can be rewritten as
$$b^T = v^T A^T $$



The minimum (least square) solution is given by the Moore-Penrose pseudo-inverse:
$$A_0^T = (v^T)^+ b^T = v,(v^T v)^{-1}, b^T$$



Which gives
$$A_0 = frac{1}{|v|^2} b,v^T$$



Note that $A_0$ is of rank $1$.



The other solutions are given by $$A = A_0 + B = frac{1}{|v|^2} b,v^T + B $$
For any matrix $B$ such that $B,v = 0$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In place of a matrix $B$ plus a constraint $(Bv=0)$, it might be easier to use an unconstrained matrix $C$ and the nullspace projector $(I-vv^+)$ to write the solution as $$A = bv^+ + C(I-vv^+)$$ where $v^+=frac{v^T}{v^Tv}quad$ This is actually the same solution since $A_0=bv^+$ and $B=C(I-vv^+)$
    $endgroup$
    – greg
    Dec 27 '18 at 15:39












  • $begingroup$
    @greg Effectively. Thank you for this precision
    $endgroup$
    – Damien
    Dec 27 '18 at 16:05











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The first column of $A$ is $A.(1,0,0,ldots,0)$, the second column is $A.(0,1,0,ldots,0)$ and so on.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well the algorithm starts with $A_0 = gamma I$ so your answer is kind of correct. But I wanted to know $A$ after the algorithm, given that I know the result $Av$ and $v$. If it helps I am referring to the algorithm called two-loop recursion of Quasi-Newton method.
    $endgroup$
    – Kart Alexius
    Jan 6 at 15:05










  • $begingroup$
    I don't know. My suggestion is that you post a new question, but this time don't forget to mention that algorithm.
    $endgroup$
    – José Carlos Santos
    Jan 6 at 15:09


















1












$begingroup$

The first column of $A$ is $A.(1,0,0,ldots,0)$, the second column is $A.(0,1,0,ldots,0)$ and so on.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well the algorithm starts with $A_0 = gamma I$ so your answer is kind of correct. But I wanted to know $A$ after the algorithm, given that I know the result $Av$ and $v$. If it helps I am referring to the algorithm called two-loop recursion of Quasi-Newton method.
    $endgroup$
    – Kart Alexius
    Jan 6 at 15:05










  • $begingroup$
    I don't know. My suggestion is that you post a new question, but this time don't forget to mention that algorithm.
    $endgroup$
    – José Carlos Santos
    Jan 6 at 15:09
















1












1








1





$begingroup$

The first column of $A$ is $A.(1,0,0,ldots,0)$, the second column is $A.(0,1,0,ldots,0)$ and so on.






share|cite|improve this answer









$endgroup$



The first column of $A$ is $A.(1,0,0,ldots,0)$, the second column is $A.(0,1,0,ldots,0)$ and so on.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 26 '18 at 11:23









José Carlos SantosJosé Carlos Santos

163k22131234




163k22131234












  • $begingroup$
    Well the algorithm starts with $A_0 = gamma I$ so your answer is kind of correct. But I wanted to know $A$ after the algorithm, given that I know the result $Av$ and $v$. If it helps I am referring to the algorithm called two-loop recursion of Quasi-Newton method.
    $endgroup$
    – Kart Alexius
    Jan 6 at 15:05










  • $begingroup$
    I don't know. My suggestion is that you post a new question, but this time don't forget to mention that algorithm.
    $endgroup$
    – José Carlos Santos
    Jan 6 at 15:09




















  • $begingroup$
    Well the algorithm starts with $A_0 = gamma I$ so your answer is kind of correct. But I wanted to know $A$ after the algorithm, given that I know the result $Av$ and $v$. If it helps I am referring to the algorithm called two-loop recursion of Quasi-Newton method.
    $endgroup$
    – Kart Alexius
    Jan 6 at 15:05










  • $begingroup$
    I don't know. My suggestion is that you post a new question, but this time don't forget to mention that algorithm.
    $endgroup$
    – José Carlos Santos
    Jan 6 at 15:09


















$begingroup$
Well the algorithm starts with $A_0 = gamma I$ so your answer is kind of correct. But I wanted to know $A$ after the algorithm, given that I know the result $Av$ and $v$. If it helps I am referring to the algorithm called two-loop recursion of Quasi-Newton method.
$endgroup$
– Kart Alexius
Jan 6 at 15:05




$begingroup$
Well the algorithm starts with $A_0 = gamma I$ so your answer is kind of correct. But I wanted to know $A$ after the algorithm, given that I know the result $Av$ and $v$. If it helps I am referring to the algorithm called two-loop recursion of Quasi-Newton method.
$endgroup$
– Kart Alexius
Jan 6 at 15:05












$begingroup$
I don't know. My suggestion is that you post a new question, but this time don't forget to mention that algorithm.
$endgroup$
– José Carlos Santos
Jan 6 at 15:09






$begingroup$
I don't know. My suggestion is that you post a new question, but this time don't forget to mention that algorithm.
$endgroup$
– José Carlos Santos
Jan 6 at 15:09













1












$begingroup$

There exists an infinity of solutions to the problem
$$b=Av$$
Where $b$ and $v$ are vectors



The problem can be rewritten as
$$b^T = v^T A^T $$



The minimum (least square) solution is given by the Moore-Penrose pseudo-inverse:
$$A_0^T = (v^T)^+ b^T = v,(v^T v)^{-1}, b^T$$



Which gives
$$A_0 = frac{1}{|v|^2} b,v^T$$



Note that $A_0$ is of rank $1$.



The other solutions are given by $$A = A_0 + B = frac{1}{|v|^2} b,v^T + B $$
For any matrix $B$ such that $B,v = 0$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In place of a matrix $B$ plus a constraint $(Bv=0)$, it might be easier to use an unconstrained matrix $C$ and the nullspace projector $(I-vv^+)$ to write the solution as $$A = bv^+ + C(I-vv^+)$$ where $v^+=frac{v^T}{v^Tv}quad$ This is actually the same solution since $A_0=bv^+$ and $B=C(I-vv^+)$
    $endgroup$
    – greg
    Dec 27 '18 at 15:39












  • $begingroup$
    @greg Effectively. Thank you for this precision
    $endgroup$
    – Damien
    Dec 27 '18 at 16:05
















1












$begingroup$

There exists an infinity of solutions to the problem
$$b=Av$$
Where $b$ and $v$ are vectors



The problem can be rewritten as
$$b^T = v^T A^T $$



The minimum (least square) solution is given by the Moore-Penrose pseudo-inverse:
$$A_0^T = (v^T)^+ b^T = v,(v^T v)^{-1}, b^T$$



Which gives
$$A_0 = frac{1}{|v|^2} b,v^T$$



Note that $A_0$ is of rank $1$.



The other solutions are given by $$A = A_0 + B = frac{1}{|v|^2} b,v^T + B $$
For any matrix $B$ such that $B,v = 0$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In place of a matrix $B$ plus a constraint $(Bv=0)$, it might be easier to use an unconstrained matrix $C$ and the nullspace projector $(I-vv^+)$ to write the solution as $$A = bv^+ + C(I-vv^+)$$ where $v^+=frac{v^T}{v^Tv}quad$ This is actually the same solution since $A_0=bv^+$ and $B=C(I-vv^+)$
    $endgroup$
    – greg
    Dec 27 '18 at 15:39












  • $begingroup$
    @greg Effectively. Thank you for this precision
    $endgroup$
    – Damien
    Dec 27 '18 at 16:05














1












1








1





$begingroup$

There exists an infinity of solutions to the problem
$$b=Av$$
Where $b$ and $v$ are vectors



The problem can be rewritten as
$$b^T = v^T A^T $$



The minimum (least square) solution is given by the Moore-Penrose pseudo-inverse:
$$A_0^T = (v^T)^+ b^T = v,(v^T v)^{-1}, b^T$$



Which gives
$$A_0 = frac{1}{|v|^2} b,v^T$$



Note that $A_0$ is of rank $1$.



The other solutions are given by $$A = A_0 + B = frac{1}{|v|^2} b,v^T + B $$
For any matrix $B$ such that $B,v = 0$






share|cite|improve this answer











$endgroup$



There exists an infinity of solutions to the problem
$$b=Av$$
Where $b$ and $v$ are vectors



The problem can be rewritten as
$$b^T = v^T A^T $$



The minimum (least square) solution is given by the Moore-Penrose pseudo-inverse:
$$A_0^T = (v^T)^+ b^T = v,(v^T v)^{-1}, b^T$$



Which gives
$$A_0 = frac{1}{|v|^2} b,v^T$$



Note that $A_0$ is of rank $1$.



The other solutions are given by $$A = A_0 + B = frac{1}{|v|^2} b,v^T + B $$
For any matrix $B$ such that $B,v = 0$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 26 '18 at 15:34

























answered Dec 26 '18 at 15:22









DamienDamien

59714




59714












  • $begingroup$
    In place of a matrix $B$ plus a constraint $(Bv=0)$, it might be easier to use an unconstrained matrix $C$ and the nullspace projector $(I-vv^+)$ to write the solution as $$A = bv^+ + C(I-vv^+)$$ where $v^+=frac{v^T}{v^Tv}quad$ This is actually the same solution since $A_0=bv^+$ and $B=C(I-vv^+)$
    $endgroup$
    – greg
    Dec 27 '18 at 15:39












  • $begingroup$
    @greg Effectively. Thank you for this precision
    $endgroup$
    – Damien
    Dec 27 '18 at 16:05


















  • $begingroup$
    In place of a matrix $B$ plus a constraint $(Bv=0)$, it might be easier to use an unconstrained matrix $C$ and the nullspace projector $(I-vv^+)$ to write the solution as $$A = bv^+ + C(I-vv^+)$$ where $v^+=frac{v^T}{v^Tv}quad$ This is actually the same solution since $A_0=bv^+$ and $B=C(I-vv^+)$
    $endgroup$
    – greg
    Dec 27 '18 at 15:39












  • $begingroup$
    @greg Effectively. Thank you for this precision
    $endgroup$
    – Damien
    Dec 27 '18 at 16:05
















$begingroup$
In place of a matrix $B$ plus a constraint $(Bv=0)$, it might be easier to use an unconstrained matrix $C$ and the nullspace projector $(I-vv^+)$ to write the solution as $$A = bv^+ + C(I-vv^+)$$ where $v^+=frac{v^T}{v^Tv}quad$ This is actually the same solution since $A_0=bv^+$ and $B=C(I-vv^+)$
$endgroup$
– greg
Dec 27 '18 at 15:39






$begingroup$
In place of a matrix $B$ plus a constraint $(Bv=0)$, it might be easier to use an unconstrained matrix $C$ and the nullspace projector $(I-vv^+)$ to write the solution as $$A = bv^+ + C(I-vv^+)$$ where $v^+=frac{v^T}{v^Tv}quad$ This is actually the same solution since $A_0=bv^+$ and $B=C(I-vv^+)$
$endgroup$
– greg
Dec 27 '18 at 15:39














$begingroup$
@greg Effectively. Thank you for this precision
$endgroup$
– Damien
Dec 27 '18 at 16:05




$begingroup$
@greg Effectively. Thank you for this precision
$endgroup$
– Damien
Dec 27 '18 at 16:05


















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