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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.

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up vote 17 down vote accepted

Yes, it does imply exactly what you've said. Remember that linear map between vector spaces of the same finite dimension is injective if and only if it is surjective if and only if it is bijective. Furthermore, it's a good exercise to carry out that such a map always takes a linearly independent set to a linearly independent set, which implies that such a map takes a basis of $L$ to a basis of $M$ (when they have the same dimension).

In this case, choose any basis $\{e_i\}$ of $L$ and then consider $\{fe_i\}$ as a basis of $M$. Then the matrix of $f$ with respect to these bases is the identity.

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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.

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