I know that there are many proofs regarding this. However, the book i'm using seems to suggest another way to do it without giving an answer. What i mean by the another way is some other proofs that do not use the fact that elementary row operation can be expressed by multiplying elementary matrices. The book says that the lemma need to be proved only when the size of identity matrix is 2 by 2. Does it have something to do with mathematical induction? That is, did the author want us to find out both induction hypothesis and induction step?
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I will take $K$ a field, $n\geq 1$ and identify matrices $M_n(K)$ with linear endomorphisms of the space $K^n$ using the canonical base $(e_1,...,e_n)$ of this space.
I think I see what this means. Let us set :
$$(E_{k,l})_{i,j}=1 \text{ if } i=k\text{ and } j=l $$
$$(E_{k,l})_{i,j}=0 \text{ otherwise} $$
Then the elementary matrices are $I_n+\lambda E_{k,l}$ with $k\neq l$, $\lambda\in K$ and $\phi$ the corresponding linear application. Now to show that such a matrix is invertible we will decompose the space :
$$K^n=Vect(e_k,e_l)\oplus Vect(e_1,...,\hat{e_k},...,\hat{e_l},...,e_n) $$
Clearly $\phi$ respect this sum and induce identity on the second space. On the first space it induces :
$$\begin{pmatrix}1&\lambda\\0&1\end{pmatrix} $$
Which is an elementary matrix which is $2$ by $2$.If I can invert this one then I can invert $\phi$ by taking the direct sum of the inverse of the $2$ by $2$ matrix on $Vect(e_k,e_l)$ and identity on the other subspace.
There is no induction in this proof.
Clément Guérin
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I think it should be $I_n+\lambda E_{kl}$, for more generality (row operations usually involve a coefficient). But the proof won't change much – Jean-Claude Arbaut Apr 14 '15 at 15:24
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@Jean-ClaudeArbaut, I was not sure about the definition of elementary matrix, I edited, thanks. – Clément Guérin Apr 14 '15 at 15:27