Solve for a matrix part of a matrix product

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I have the equation

\mathbf{A}=\mathbf{W}_1\mathbf{T}\mathbf{W}_2 that represent some measurement setup I have at the uni. lab. The matrices are given as

\mathbf{A}= \begin{bmatrix}a1 & a2\\a2 & a1\end{bmatrix}\,\mathbf{W}_1=\begin{bmatrix}w_{11} & w_{12}\\w_{21} & w_{22}\end{bmatrix}\,\mathbf{T}=\begin{bmatrix}t_1 & t_2 \\ t_2 & t_1\end{bmatrix}\,\mathbf{W}_2=\begin{bmatrix}w_{22} & w_{21}\\ w_{12} & w_{11}\end{bmatrix}

The thing is that I can measure the data representing A and T, however, measuring W by itself is tricky, so re-arranging the setup to match the equation above is the easiest way, practically speaking. But it is W that represents the data that I want, so I must isolate it in some way. From the above equation I can't come up with any straight forward way; I've tried literally all tricks I learned in when I had linear algebra a few years back, but no can't do. I know that, most likely, there are some tricks that I've forgotten, so maybe someone in here can lead me in the right direction?

Listing up my different attempts would take too much time for writing the TeX code, but in short I can say that I always end up with a "solution" where I have unknown matrices on both sides of a known matrix, with other words back to start, but with an uglier expression.

Any help is appreciated!
 
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i'm not sure i understand the problem right, but does it help to notice you can write
<br /> \mathbf{W}_2<br /> = \begin{bmatrix}0&amp; 1\\1 &amp; 0\end{bmatrix}<br /> \begin{bmatrix}w_{11 } &amp; w_{12}\\w_{21} &amp; w_{22}\end{bmatrix}<br /> \begin{bmatrix}0&amp; 1\\1 &amp; 0\end{bmatrix}<br /> = <br /> \mathbf{P}\mathbf{W}_1\mathbf{P}<br /> <br />

with
<br /> \mathbf{P}= \begin{bmatrix}0&amp; 1\\1 &amp; 0\end{bmatrix}<br />

then at least your problem is now maybe a little clearer
\mathbf{A}=\mathbf{W}\mathbf{T}\mathbf{P}\mathbf{W}\mathbf{P}

and as P is its own inverse you get to
\mathbf{P}\mathbf{A}=\mathbf{W}\mathbf{T}\mathbf{P}\mathbf{W}
 
I see that my formulation became a bit unclear. Your alternative representation is indeed better - didn't think of that.

However, even from \mathbf{P}\mathbf{A}=\mathbf{W}\mathbf{T}\mathbf{P}\mathbf{W} I will still end up with expressions that are quite similar to the original expression. I.e. unknown * known * unknown = known, or unknown*known = known*unknown, so no matter what I do (of the tricks that I know about), I can't isolate the unknown. However, even a solution like \mathbf{W}^T\mathbf{W}=\mathbf{X} where \mathbf{X} is some other matrix product would do (or something similar - as long as there's only \mathbf{W}s in the expression at one side of the equality sign), as I always can make assumptions on the data and then see if they hold.

Anyways, thanks for the help making the problem clearer. Think I just have to chew a bit more on this one.
 
yeah i think i understand your problem a bit better, the only thing i could think of is maybe to reformulate the problem, instead consider the vector
\mathbf{w}=\begin{bmatrix}w_1 \\ w_{2}\\ w_{3} \\ w_{4}\end{bmatrix}

then see if you can re-write the equation to be something more like a quadratic form
a = \mathbf{w}^T \mathbf{T}&#039; \mathbf{w}
where T' is now a 4x4 matrix

from there you can try and digonalise T and maybe solve...
 
Hm, I'm not sure if I get this 100%. How should I express/extend the \mathbf{T}&#039; matrix before trying to diagonalize it?

I also see know that I should have posted this in the Linear & Abstract algebra forum instead, as this isn't a homework/text-book question ;)
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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