Q:

Identity-Matrix Normal form of a linear transformation

I'm confused about the following.
Let $\phi:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be a linear isomorphism and let $\nu=(v_1,...v_n)$ be a basis for $\mathbb{R}^n$ and $\omega=(w_1,...,w_n)$ the basis for $\mathbb{R}^n$ given by $w_j:=\phi(v_j)$. Now if I look at the matrix which represent $\phi$ with respect to the basis $\nu$ and $\omega$ I get that this matrix is the identity matrix. But this can't be true. Where is my mistake?
(If this would be true, then any isomorphism would have determinant 1, since the determinant of a linear map is independent of the chosen repr. matrix)
On the other hand I read that every linear mapping $A: 
\mathbb{R}^n\rightarrow \mathbb{R}^m$ has a normal form 
$
\text{diag}(1,...,1,0,...,0)
$,
where there are r=rank(A) many 1's 
. Is this even true then? And if so, what would be a proof?(when reading this I thought that the proof is to choose a basis as above and then add linearly independ vectors of the kernel of A, but I guess this is wrong)

A:

There is no mistake.
What probably confuses you is that well... what's the point of having other non singular matrices if you can express isomorphisms via the identity matrix?
The problem is that those matrices express isomorphisms for bases that are non correspondent one by one under that isomorphism.
Summarizing, the "bizarre" thing about this isomorphism is that you have selected a very special base for the image: a base composed by the images of the base of the dominion.

