# How can I obtain the generic matrix T such that hat(X)=TX with: hat(X)=[[a,0],[c,0],[0,b],[0,d]] and X=[[a,c],[b,d]]

How can I obtain the generic matrix T such that $\stackrel{^}{X}=TX$ with:
$\stackrel{^}{X}=\left[\begin{array}{cc}a& 0\\ 0& b\\ c& 0\\ 0& d\end{array}\right]$
and
$X=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
I did several try without obtaining the result I needed.
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Illuddybopylu
Is X is invertible? Then just multilpy ${X}^{-1}$ at right of both side of equation.Then T will be $\stackrel{^}{X}{X}^{-1}$

Staffangz
There is no such T that's independent of the entries of X.If there were such T, then consider the effect of swapping the columns of X,
${X}^{\prime }=\left[\begin{array}{cc}b& a\\ d& c\end{array}\right]=X\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
On one hand, T was assumed to be generic and would transform X′ to the same format:
$T{X}^{\prime }=\left[\begin{array}{cc}b& 0\\ 0& a\\ d& 0\\ 0& c\end{array}\right]$
On the other hand, TX′ should be like TX but with swapped columns:
$T{X}^{\prime }=TX\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}a& 0\\ 0& b\\ c& 0\\ 0& d\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0& a\\ b& 0\\ 0& c\\ d& 0\end{array}\right]$
And note how the zero entries are at different positions.