Restriction on choosing a basis for $T:V\to V$ At about 49:10 in this video, the lecturer discusses linear endomorphisms
$$T:V \to V $$
and how a good choice of basis for $V$ could simplify the matrix $[T]$ representing this map. The lecturer says that one cannot use two different bases of $V$, one for the domain and one for the image. Why is that the case?
Thanks.
 A: You certainly can allow yourself to use two different bases, but this completely changes the nature of the whole problem. Usually one probably discusses maps $T:V\to W$ before discussing maps $V\to V$. If you allow yourself to use two different bases, then what you get is just the results for maps $V\to W$ specialized to the case $W=V$: Then only invariant is the rank, and if the rank equals $r$, then you can choose bases in such a way that $A=(a_{ij})$ with $a_{11}=\dots=a_{rr}=1$ and all other entries $0$. (Just take a basis for the kernel of $T$ which has $n-r$ elements, extend it to a basis of $V$ and order it in such a way that the $r$ additional elements come first. Then take the images of these $r$ elements under $T$ and extend to a basis of $V$.)
Clearly, this is a much more crude classification of linear maps $V\to V$ that you actually want. For example, every linear isomorphism can be represented by the identity matrix if you are willing to use two different bases. You cannot even distinguish different multiples of the identity map if you don't restrict to using the same basis both times. 
