Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following problem:

Given two 2D real positive-definite symmetric matrices $M_1$ and $M_2$, find a matrix $T$ such that $$ M_2=TM_1T^t$$

Clearly, the solution is not unique, but I don't care too much about that. All I need is a well-defined solution - something like a principal branch.

To be concrete, let's say I have some $T_{input}$, and I generate $M_1$ randomly and calculate $M_2=TM_1T^t$. I want a function $T_{output}=f(M_1,M_2)$ that will give me the same result (which is not necessarily $T_{input}$) regardless of my choice of $M_1$.

My thoughts: For a symmetric positive definite matrix $A$, the principal square root of the matrix is uniquely defined, ans is also symmetric. One solution of of the problem is thus $$T=\sqrt{M_2}\left(\sqrt{M_1}\right)^{-1}\ .$$ It is easily seen that this is a solution. Since $\sqrt{M_1}\sqrt{M_1}=M_1$, we have $$\left(\sqrt{M_1}\right)^{-1}M_1\left(\sqrt{M_1}\right)^{-1}=I\ ,$$ and thus $$ \begin{align} TM_1T^t&=\sqrt{M_2}\left(\sqrt{M_1}\right)^{-1}\ M_1\ \left(\sqrt{M_1}\right)^{-1} \sqrt{M_2}\\ &=\sqrt{M_2}\ I \ \sqrt{M_2}=M_2\\ \end{align} $$

But this solution depends on the choice of $M_1$. Any suggestions will be greatly appreciated.

share|improve this question
    
This was asked at MO... –  J. M. Jul 12 '12 at 11:57
    
@J.M.: As I understand the question, that MO question is a different question, but I might be misunderstanding. –  joriki Jul 12 '12 at 13:17
    
They are the same to me, @joriki. The $\mathbf A$ and $\mathbf B$ matrices in the MO question would correspond to the $\mathbf M_1$ and $\mathbf M_2$ matrices in OP's question. –  J. M. Jul 12 '12 at 13:25
    
@J.M.: I don't see the requirement "I want a function $T_{output}=f(M_1,M_2)$ that will give me the same result regardless of my choice of $M_1$" in the MO question. I'm not entirely sure I understand that requirement correctly, but under my understanding (see my answer), this is a crucial constraint that makes a solution impossible. –  joriki Jul 12 '12 at 13:31
    
@joriki, you're right; that certainly throws a wrench into things... –  J. M. Jul 12 '12 at 13:34
add comment

1 Answer 1

up vote 2 down vote accepted

Let me restate the question, because I'm not entirely sure I understand it correctly. As I understand it, you want a function that takes $M$ and $TMT^\top$ and returns a matrix $S$ that depends only on $T$ and not on $M$, such that $SMS^\top=TMT^\top$.

This is impossible. Given $T$, since $SMS^\top=TMT^\top$ for all positive-definite symmetric $M$, we have

$$T^{-1}SMS^\top T^\top{}^{-1}=(T^{-1}S)M(T^{-1}S)^\top=M$$

for all positive-definite symmetric $M$. In particular, for $M=I$, we have $(T^{-1}S)(T^{-1}S)^\top=I$, so $T^{-1}S$ is orthogonal. Thus we have

$$(T^{-1}S)M=M(T^{-1}S)$$

for all positive-definite symmetric $M$. But the only matrices that commute with all positive-definite symmetric matrices are multiples of the identity. Thus $S=\pm T$. So your function would have to return $T$ (up to a sign), which it can't, since different $T$s can lead to the same inputs.

share|improve this answer
    
You correctly understood my question, and answered me. I don't like the answer though, because it means that I need to work harder now in my research... :( Thanks! –  yohBS Jul 12 '12 at 13:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.