How to answer:
Singular Value Decomposition is a rank revealing matrix factorization?
I have been checking several numerical linear algebra books, most of them define rank revealing factorization as and specific way to write the QR decomposition. But I have not found how to know if any arbitrary factorization is rank revealing.
The books in wich I have search definition are: https://www.scribd.com/document/323104852/LAFF-NLA https://epubs.siam.org/doi/abs/10.1137/1.9781611971484
Please advise
A "rank revealing factorization" or "rank factorization" of a rank $r$ matrix $A$ refers to any factorization of the form $A = CF$ in which $C$ has size $m \times r$ and $F$ has size $r \times n$. Note that this factorization is highly non-unique. In particular, if $A = CF$ is one factorization and $S$ is any invertible $r \times r$ matrix, then $A = (CS)(S^{-1}F)$ is another such factorization.
The factorization is called "rank revealing" because it "reveals the rank" in the following sense: an $m\times n$ matrix $A$ can be written in the form $A = CF$ with $C,F$ of sizes $m \times r$ and $r \times n$ if and only if $A$ has rank less than or equal to $r$.
I suspect that the numerical linear algebra textbooks are referring to the various ways in which a rank factorization can be obtained from other decompositions. For example, we can obtain a rank-revealing factorization similarly to how one might obtain a "thin" QR factorization. Note that in the full QR factorization of a matrix $A$ with rank $r$, we can write $$ A = QR = \pmatrix{Q_1 & Q_2} \pmatrix{R_1\\ 0}, $$ where $Q_1$ has $r$ columns (that form an orthonormal basis for the column space of $A$), and $R_1$ is $r \times n$ and upper triangular. With block-matrix multiplication, we see that $$ A = \pmatrix{Q_1 & Q_2} \pmatrix{R_1\\ 0} = Q_1 R_1 + Q_2 0 = Q_1 R_1. $$ We see that $A = Q_1 R_1$ is a rank-revealing factorization because $Q_1,R_1$ have sizes $m \times r$ and $r \times n$ respectively.
Similarly, a compact SVD can be thought of as a rank-revealing factorization in that we have $A = (U_r \Sigma_r)V_r^*$.