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A = B * C, how do I solve for B

0

$begingroup$

$$A = B C$$

where $A$'s dimension = $n$ x $1$, $B$ 's dimension = $n$ x $n$, C 's dimension = $A$ = $n$ x $1$

I know A and C.

How do I solve for B?

Attempt: I was thinking about multiply by the inverse of C. But C is non invertiable because its dimension nx1.

---

Context: statweb.stanford.edu/~tibs/sta305files/Rudyregularization.pdf (page 26, I am looking to solve for the Smoother Matrix S).

matrices problem-solving matrix-equations matrix-decomposition

share|cite|improve this question

edited Mar 27 at 18:37

wrek

asked Mar 27 at 18:23

wrekwrek

124214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is generally not a unique solution: there will be multiple $B$'s satisfying this equation. Are you looking to find just one of them? If so, how about making $B$ diagonal, with suitable diagonal entries?
  $endgroup$
  – avs
  Mar 27 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am trying to find the smoother matrix as indicated here: statweb.stanford.edu/~tibs/sta305files/Rudyregularization.pdf (page 26)
  $endgroup$
  – wrek
  Mar 27 at 18:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @avs the process you mentioned is only valid if all the entries of $C$ are non zero.
  $endgroup$
  – Dbchatto67
  Mar 27 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dbchatto67, why? If $C$ has a zero entry, just make the corresponding diagonal entry in $B$ zero. $B$ only has to map the specific vector $A$ to the specific vector $C$, as I understood correctly.
  $endgroup$
  – avs
  Mar 27 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @wrek, the context in your comment is important: you should included it in your question. Otherwise, people might give your question negative votes.
  $endgroup$
  – avs
  Mar 27 at 18:34
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

$$A = B C$$

where $A$'s dimension = $n$ x $1$, $B$ 's dimension = $n$ x $n$, C 's dimension = $A$ = $n$ x $1$

I know A and C.

How do I solve for B?

Attempt: I was thinking about multiply by the inverse of C. But C is non invertiable because its dimension nx1.

---

Context: statweb.stanford.edu/~tibs/sta305files/Rudyregularization.pdf (page 26, I am looking to solve for the Smoother Matrix S).

matrices problem-solving matrix-equations matrix-decomposition

share|cite|improve this question

edited Mar 27 at 18:37

wrek

asked Mar 27 at 18:23

wrekwrek

124214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is generally not a unique solution: there will be multiple $B$'s satisfying this equation. Are you looking to find just one of them? If so, how about making $B$ diagonal, with suitable diagonal entries?
  $endgroup$
  – avs
  Mar 27 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am trying to find the smoother matrix as indicated here: statweb.stanford.edu/~tibs/sta305files/Rudyregularization.pdf (page 26)
  $endgroup$
  – wrek
  Mar 27 at 18:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @avs the process you mentioned is only valid if all the entries of $C$ are non zero.
  $endgroup$
  – Dbchatto67
  Mar 27 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dbchatto67, why? If $C$ has a zero entry, just make the corresponding diagonal entry in $B$ zero. $B$ only has to map the specific vector $A$ to the specific vector $C$, as I understood correctly.
  $endgroup$
  – avs
  Mar 27 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @wrek, the context in your comment is important: you should included it in your question. Otherwise, people might give your question negative votes.
  $endgroup$
  – avs
  Mar 27 at 18:34
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

$$A = B C$$

where $A$'s dimension = $n$ x $1$, $B$ 's dimension = $n$ x $n$, C 's dimension = $A$ = $n$ x $1$

I know A and C.

How do I solve for B?

Attempt: I was thinking about multiply by the inverse of C. But C is non invertiable because its dimension nx1.

---

Context: statweb.stanford.edu/~tibs/sta305files/Rudyregularization.pdf (page 26, I am looking to solve for the Smoother Matrix S).

matrices problem-solving matrix-equations matrix-decomposition

share|cite|improve this question

edited Mar 27 at 18:37

wrek

asked Mar 27 at 18:23

wrekwrek

124214

$endgroup$

share|cite|improve this question

edited Mar 27 at 18:37

wrek

asked Mar 27 at 18:23

wrekwrek

124214

share|cite|improve this question

edited Mar 27 at 18:37

wrek

asked Mar 27 at 18:23

wrekwrek

124214

asked Mar 27 at 18:23

wrekwrek

124214

asked Mar 27 at 18:23

wrekwrek

124214

1 Answer
1

active

oldest

votes

0

$begingroup$

You can't solve for a unique $B$. For example let $A=kC$, therefore$$BC=kC$$then the answer $B$ to the above equation includes all matrices that have an eigenvalue $k$ and a corresponding eigenvector $C$.

Note that if $B$ is not symmetric, then its trace is not unique. For example if$$A=C=beginbmatrix1\1endbmatrix$$and$$B=beginbmatrixa&1-a\b&1-bendbmatrix$$then the equation $A=BC$ is satisfied for any $a,bin Bbb R$, but if $B$ is symmetric, we from $A=BC$ we can write $$C^TA=C^TBC$$therefore$$tr(C^TA)=tr(C^TBC)=tr(BC^TC)=C^TCcdot tr(B)$$hence$$tr(B)=A^TCover C^TC$$where I exploited the properties of trace.

share|cite|improve this answer

edited Mar 27 at 18:53

answered Mar 27 at 18:37

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am looking for the trace of $B$
  $endgroup$
  – wrek
  Mar 27 at 18:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @wrek: I think you can always get the trace of $B$ to be $0$ (assuming $nge 2$).
  $endgroup$
  – Henning Makholm
  Mar 27 at 18:42
  
  

  
  
  
  

add a comment | 

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0

$begingroup$

You can't solve for a unique $B$. For example let $A=kC$, therefore$$BC=kC$$then the answer $B$ to the above equation includes all matrices that have an eigenvalue $k$ and a corresponding eigenvector $C$.

Note that if $B$ is not symmetric, then its trace is not unique. For example if$$A=C=beginbmatrix1\1endbmatrix$$and$$B=beginbmatrixa&1-a\b&1-bendbmatrix$$then the equation $A=BC$ is satisfied for any $a,bin Bbb R$, but if $B$ is symmetric, we from $A=BC$ we can write $$C^TA=C^TBC$$therefore$$tr(C^TA)=tr(C^TBC)=tr(BC^TC)=C^TCcdot tr(B)$$hence$$tr(B)=A^TCover C^TC$$where I exploited the properties of trace.

share|cite|improve this answer

edited Mar 27 at 18:53

answered Mar 27 at 18:37

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am looking for the trace of $B$
  $endgroup$
  – wrek
  Mar 27 at 18:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @wrek: I think you can always get the trace of $B$ to be $0$ (assuming $nge 2$).
  $endgroup$
  – Henning Makholm
  Mar 27 at 18:42
  
  

  
  
  
  

add a comment | 

0

$begingroup$

You can't solve for a unique $B$. For example let $A=kC$, therefore$$BC=kC$$then the answer $B$ to the above equation includes all matrices that have an eigenvalue $k$ and a corresponding eigenvector $C$.

Note that if $B$ is not symmetric, then its trace is not unique. For example if$$A=C=beginbmatrix1\1endbmatrix$$and$$B=beginbmatrixa&1-a\b&1-bendbmatrix$$then the equation $A=BC$ is satisfied for any $a,bin Bbb R$, but if $B$ is symmetric, we from $A=BC$ we can write $$C^TA=C^TBC$$therefore$$tr(C^TA)=tr(C^TBC)=tr(BC^TC)=C^TCcdot tr(B)$$hence$$tr(B)=A^TCover C^TC$$where I exploited the properties of trace.

share|cite|improve this answer

edited Mar 27 at 18:53

answered Mar 27 at 18:37

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am looking for the trace of $B$
  $endgroup$
  – wrek
  Mar 27 at 18:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @wrek: I think you can always get the trace of $B$ to be $0$ (assuming $nge 2$).
  $endgroup$
  – Henning Makholm
  Mar 27 at 18:42
  
  

  
  
  
  

add a comment | 

$begingroup$

You can't solve for a unique $B$. For example let $A=kC$, therefore$$BC=kC$$then the answer $B$ to the above equation includes all matrices that have an eigenvalue $k$ and a corresponding eigenvector $C$.

Note that if $B$ is not symmetric, then its trace is not unique. For example if$$A=C=beginbmatrix1\1endbmatrix$$and$$B=beginbmatrixa&1-a\b&1-bendbmatrix$$then the equation $A=BC$ is satisfied for any $a,bin Bbb R$, but if $B$ is symmetric, we from $A=BC$ we can write $$C^TA=C^TBC$$therefore$$tr(C^TA)=tr(C^TBC)=tr(BC^TC)=C^TCcdot tr(B)$$hence$$tr(B)=A^TCover C^TC$$where I exploited the properties of trace.

share|cite|improve this answer

edited Mar 27 at 18:53

answered Mar 27 at 18:37

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

share|cite|improve this answer

edited Mar 27 at 18:53

answered Mar 27 at 18:37

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 27 at 18:37

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 27 at 18:37

Mostafa AyazMostafa Ayaz

18.2k31040