<p>If your error norm is the Frobenius distance (that is, $m(X, U) = \|X - U \|_F$, then there is a closed form for your problem. $A = V^+ U$, where $V^+$ is the <a href="http://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse">Moore-Penrose pseudoinverse</a> of $V$ (this is a generalized inverse for non-square matrices).</p>

CC BY-SA 3.0
2012-03-21T21:02:55.873Z
Thanks. This is great. I'm curious whether there is a closed form solution if (1) we take $m$ to be the Schatten $p$-norm of $X - U$ for any $p$ (not just the Frobenius, where $p = 2$); or (2) we take $m$ to be the 2-norm--i.e. the spectral radius of $X - U$. If there is a solution for (2), I wonder whether it is general for a matrix norm induced by a vector $p$-norm for any $p$.
aaronsteven
CC BY-SA 3.0
2012-03-21T23:21:14.880Z
It's a good question. Not sure I know the answer :)
Suresh Venkat
CC BY-SA 3.0
2012-03-17T18:53:15.153Z
2012-03-17T19:05:02.843Z
2012-03-17T19:05:02.843Z
Suresh Venkat
<p>I have two matrices $U$ and $V$.  $U$ is $n \times n$ and $V$ is $n \times m$. (Both are empirical results of an experiment.) I would like to find a linear transformation $A$, $m \times n$, such that $VA = \hat{U}$, where $\hat{U} = \underset{X}{\arg \min}$ $m(X, U)$ for some metric $m$.</p>

<p>Is there a closed form or iterative solution to this problem?</p>

CC BY-SA 3.0
2012-03-17T18:26:27.520Z
What is the argmin taken over? If it is taken over all n x n matrices, then $\hat{U} = U$, so presumably you are imagining taking the argmin over a different set. What this set is and what  the metric is seem potentially crucial to an answer.
Joshua Grochow
CC BY-SA 3.0
2012-03-17T18:28:10.093Z
Your formulation implies that any linear transformation A is acceptable to you as long as the product VA is close to U, but are you sure of this?
Tsuyoshi Ito
CC BY-SA 3.0
2012-03-17T18:44:36.647Z
@JoshuaGrochow even if the argmin is over all $n\times n$ matrices, it's not clear that it's possible to get $U$, so the question is still well defined (the implicit restriction on X is that it's the image of f(A) = VA)
Suresh Venkat
CC BY-SA 3.0
2012-03-17T19:36:13.050Z
@Suresh: That is what I first thought, but it is not correct.  Note that if the argmin is really over all n×n matrices, then it has nothing to do with the product VA!
Tsuyoshi Ito
CC BY-SA 3.0
2012-03-17T20:41:35.533Z
Right. So X has to be restricted to matrices that can be formed as VA for a fixed V
Suresh Venkat
<linear-algebra><matrices><na.numerical-analysis>
Finding mapping between two spatial representations of the same objects
aaronsteven
# Finding mapping between two spatial representations of the same objects

(Asked by: username_0 on 2012-03-17)

I have two matrices $U$ and $V$. $U$ is $n \times n$ and $V$ is $n \times m$. (Both are empirical results of an experiment.) I would like to find a linear transformation $A$, $m \times n$, such that $VA = \hat{U}$, where $\hat{U} = \underset{X}{\arg \min}$ $m(X, U)$ for some metric $m$.

Is there a closed form or iterative solution to this problem?

**Tags:** linear-algebra, matrices, na.numerical-analysis

> **Comments:**
> **username_1**: What is the argmin taken over? If it is taken over all n x n matrices, then $\hat{U} = U$, so presumably you are imagining taking the argmin over a different set. What this set is and what  the metric is seem potentially crucial to an answer.  
> **username_2**: Your formulation implies that any linear transformation A is acceptable to you as long as the product VA is close to U, but are you sure of this?  
> **username_3**: @JoshuaGrochow even if the argmin is over all $n\times n$ matrices, it's not clear that it's possible to get $U$, so the question is still well defined (the implicit restriction on X is that it's the image of f(A) = VA)  
> **username_2**: @Suresh: That is what I first thought, but it is not correct.  Note that if the argmin is really over all n×n matrices, then it has nothing to do with the product VA!  
> **username_3**: Right. So X has to be restricted to matrices that can be formed as VA for a fixed V  

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## 1 answers

### Answer by username_3 on 2012-03-17. Score: 6 | Accepted answer

If your error norm is the Frobenius distance (that is, $m(X, U) = \|X - U \|\_F$, then there is a closed form for your problem. $A = V^+ U$, where $V^+$ is the [Moore-Penrose pseudoinverse](http://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse) of $V$ (this is a generalized inverse for non-square matrices).

> **Comments:**
> **username_0**: Thanks. This is great. I'm curious whether there is a closed form solution if (1) we take $m$ to be the Schatten $p$-norm of $X - U$ for any $p$ (not just the Frobenius, where $p = 2$); or (2) we take $m$ to be the 2-norm--i.e. the spectral radius of $X - U$. If there is a solution for (2), I wonder whether it is general for a matrix norm induced by a vector $p$-norm for any $p$.  
> **username_3**: It's a good question. Not sure I know the answer :)  

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