In linear algebra, we know that a system of equations $AX=b$ can be easily solved if $A$ is found to be of diagonal nature. If however $A$ is not diagonal but can be changed into a diagonal form by using the change of variables how can it be shown that the original system $AX=b$ and the new system $DY=z$ has indeed the same set of solutions?
Here, $X=Py$ and $Y=P^{-1}X$. Also $b=Pz$. So, $P^{-1}APY=z$ or $DY=z$
For example, if $AX=b$ is consistent in nature, then the transformed system $DY=z$ is also found to be consistent in nature. Likewise, if the original system is inconsistent, then $DY=z$ is also found to be inconsistent.
My Question is this : Is there any validity or proof for these assumptions ?
Thanks in advance guys! If this is a very basic question consider closing it but do post the link to the relevant solution.
In the square case, everything comes down to eigenvalues. In this case, suppose $A$ is similar to $D$ and $D$ is diagonal. Then:
When the systems are consistent, you get the correspondence of the solutions by undoing the change of variable: if $z=P^{-1}b$ and $DY=z$ then $A(PY)=b$.
In the case where $A$ is $m \times n$ with $m>n$, you can do the same thing by considering the least squares problem, because any solution to the original problem is also a solution to the least squares problem.
I'm not sure how I would handle the case $m<n$.