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I am trying to solve chain matrices multiplication. Where I encounter a problem, which sees like dead end to me. How further I can decompose $X^{-1} . Y . X = Z$ or simplify it to find matrix $X$ from this multiplication.

Where matrices $X,Y$and $Z$ are 2x2 non-singular matrices.

Any help will be appreciated.

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Intuitively, it represents a basis change. The matrix $Y$ represented in another basis (which is represented by $X$, is $Z$. This makes sense only when $X,Y,Z$ are square matrixes. –  Andrey Dec 17 '12 at 11:24
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2 Answers

Note that your equation is quadratic in $X$ and therefore the solution is not unique, and not even guaranteed to exist. For example, if $X$ is a solution, then so is $-X$. Depending on the structure of $Y$ and $Z$, you might have more.

Also, the matrix $Z$ is similar to the matrix $Y$. That is, they both represent the same linear transformation but in different bases. Therefore, your equation does not have a solution if, $\operatorname{tr}(A)\ne\operatorname{tr}(B)$ or $\operatorname{det}(A)\ne\operatorname{det}(B)$.

Otherwise there exist a solution, but as I said, it is not unique. You can easily find it by solving, enrtywise, the 4 equations.

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Thanks. In my case, $det(A)=det(B)$ but trivial solution is always zero. But when I substitute values(what they should be), it results correct. Here question is how to reach this unique solution? –  Salman Dec 17 '12 at 12:10
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Rewriting your equation as $YX = XZ$ you get a special version of a Sylvester equation. An easy solution is to vectorize this equation using the Kronecker product

$$\mathrm{vec}(AYB) = (B^T \oplus A)\mathrm{vec}(Y)$$

(where $\oplus$ shall denote the Kronecker product!). Then,

$$\mathrm{vec}(YX) = (I \oplus Y) \mathrm{vec}(X) = (Z^T \oplus I) \mathrm{vec}(X) = \mathrm{vec}(XZ)$$

and you only have to solve the simple linear equation

$$((I \oplus Y) - (Z^T \oplus I)) \mathrm{vec}(X) = 0.$$

Cheers, Wieland

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Thanks. I try to go on by this method. –  Salman Dec 17 '12 at 12:13
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