So I am reading (or at least trying to get there) about spectral theory and the diagonalization problem.
To begin one considers how linear maps on finite dim spaces compare and relate. I am reading that of linear maps $T_i: V \to W $, they can be compared via
$$T_2U_2 = U_1T_1$$ for some invertible map $U_1: V \to W$ and $U_2: W \to V.$ This formulation suggests a square commutative diagram. IN other words, linear maps differed by an invertible map leads to the study of diagoalization.
Now my question is, why can't the relation be $T_1 = T_2 U_1$? And to get back $T_1$ we take the inverse of $U_1$.
Consider $T_1,T_2:\mathbb R^2\to\mathbb R^2$ defined by \begin{align} T_1(x,y) &= (x,0)\\ T_2(x,y) &= (0,y), \end{align} with $x$ and $y$ nonzero. An invertible map $U_1:\mathbb R^2\to\mathbb R^2$ is of the form $$U_1(x,y) = (ax+by, cx+dy), $$ with $ad-bc\ne0$. So for $(x,y)\in\mathbb R^2$ we have $$T_2U_1(x,y) = T_2(ax+by,cx+dy) = (0, cx+dy), $$ which shows that $T_2U_1\ne T_1$.