TITLE: This theorem about matrices of linear maps doesn't look correct.
QUESTION [9 upvotes]: Consider the following theorem:
Theorem. Let $f\colon L\to M$ be a linear mapping of finite-dimensional vector spaces. Then there exist bases in $L$ and $M$ and a natural number $r$ such that the matrix of $f$ in these bases has the form $(a_{ij})$, where $a_{ii}=1$ for $1\leq i\leq r$ and $a_{ij}=0$ for the other values of $i,j$. Furthermore, $r$ is the rank of $f$.
This theorem doesn't make much sense to me. Doesn't it imply that, for example, if $L$ and $M$ have the same dimension, every injective linear map can be represented by the identity matrix in some basis? This looks weird.
Can you comment on this? It is a theorem in Section 8, Chapter 1 of Kostrikin and Manin's book "Linear Algebra and Geometry".
Actually, it is not copied word-by-word, but I think that I wrote exactly what they meant.

REPLY [15 votes]: Perhaps you're confused because you know that if you were to have an injective linear map $g:L\rightarrow L$ then it wouldn't necessarily be true that there was a basis of $L$ so that the matrix of $g$ in this basis was the identity.
But the stated theorem talks about an injective linear map $f:L\rightarrow M$, between two different vector spaces. So we are considering picking a basis of $L$ and (completely separately) picking a basis of $M$. This gives us much more freedom, indeed enough to make the theorem true.