TITLE: Why isn't removing zero rows an elementary operation?
QUESTION [5 upvotes]: My prof taught us that during Gaussian Elimination, we can perform three elementary operations to transform the matrix:
1) Multiple both sides of a row by a non-zero constant
2) Add or subtract rows
3) Interchanging rows
In addition to those, why isn't removing zero rows an elementary operation? It doesn't affect the system in any way. Define zero rows to be a row with no leading variables.
For example isn't $\begin{bmatrix}a & b & k\\c & d & m\end{bmatrix} \rightarrow 
\begin{bmatrix}a & b & k\\c & d & m\\0 & 0 & 0\end{bmatrix}$

REPLY [2 votes]: Doing an elementary row operation on the matrix $A$ can be seen as multiplying $A$ by a suitable invertible matrix, call it $E_1$. So the whole process leads to writing
$$
U=E_{k}\dots E_{2}E_{1}A
$$
where $E_j$ $(j=1,2,\dots,k)$ correspond to the elementary row operation we performed and $U$ is a row reduced matrix (for instance, the row reduced echelon form). Since each $E_j$ is invertible, we can write
$$
A=FU
$$
where $F=E_1^{-1}E_2^{-1}\dots E_{k}^{-1}$ and, if one performs all row swaps at the start, the matrix $F$ can be written even without any computation.
Of course no operation of this kind changes the shape of the matrices we successively get and the final matrix $U$ will have zero or more “zero rows” at the bottom.
At this point we can remove those “zero rows” from $U$; if there are $l$ of them, we can also remove the $l$ rightmost columns of $F$. Calling $F_0$ and $U_0$ the matrices so obtained we still have
$$
A=F_0U_0
$$
and both $F_0$ and $U_0$ have their rank equal to the rank of $A$. This is called a full rank decomposition, because those matrices have the maximum rank they possibly have, based on their shape; say that $A$ is $m\times n$ and its rank is $r$. Then $F_0$ will be $m\times r$ $(r\le m)$ and $U_0$ will be $r\times n$ ($r\le n$).
In particular $F_0^HF_0$ ($H$ denotes the hermitian transpose, the simple transpose when the matrices are over the reals) is $r\times r$ invertible and
$$
(F_0^HF_0)^{-1}F_0^H
$$
not only is a left inverse of $F_0$ but is its Moore-Penrose pseudoinverse $F_0^+$. Similarly, $U_0U_0^H$ is invertible and $U_0^H(U_0U_0^H)^{-1}=U_0^+$. Moreover the Moore-Penrose pseudoinverse of $A$ is
$$
A^+=U_0^+F_0^+
$$
Indeed, any time we find a full rank decomposition $A=BC$, we can write $A^+=C^+B^+$ (and the pseudoinverse of $B$ and $C$ are computed as above).
The Moore-Penrose pseudoinverse is useful in computing the least squares solutions of $Ax=b$ and this is an efficient algorithm for finding it.