Reduction to diagonal form by transvections

Remarks

Multiplying by a transvection on the left adds a multiple of one row to another row; on the right it adds a multiple of one column to another column. The proof is an induction on $n$. If the last row or the last column of $M$ has a nonzero entry, one row or column addition puts a nonzero entry at position $(n, n)$. Further additions of multiples of row $n$ and column $n$ then clear the rest of the last row and the last column. The upper left block of size $n - 1$ is treated by induction. Example: $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ becomes $\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$ after adding column $2$ to column $1$, and then $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ after subtracting column $1$ from column $2$. Each transvection has determinant $1$, so $\det M = \det D$. A refinement of the argument shows that an invertible $M$ is a product of transvections and the matrix $\operatorname{diag}(1, \dots, 1, \det M)$. Hence the transvections generate the special linear group $\mathrm{SL}_n(F)$.

Going deeper

For a student

This is a sharper form of Gaussian elimination: elementary matrices of the row-addition type already suffice, as long as both row and column operations are allowed. The theorem is the main step in the proof that $\mathrm{SL}_n(F)$ is generated by transvections. It is also a first step towards the Smith normal form, where the field is replaced by a principal ideal domain and more care is needed.