求证：可逆的行最简形矩阵必为单位矩阵（禁用行列式、秩）
Alright, let's tackle this step by step—no determinants or rank allowed, just the definition of a row-reduced echelon form (RREF) matrix and what it means for a matrix to be invertible.
First, let's recap the core properties of an n×n row-reduced echelon form (RREF) matrix:
- Any all-zero rows are positioned at the bottom of the matrix.
- The first non-zero element (called a pivot) in each non-zero row is 1.
- All elements above and below a pivot are 0.
- Pivots in successive rows lie in columns with strictly increasing indices.
Now, suppose ( R ) is an n×n RREF matrix that is invertible. Let's prove ( R = I ) (the identity matrix):
R cannot contain any all-zero rows
Suppose for contradiction that ( R ) has at least one all-zero row. Take the vector ( \mathbf{x} = (0, 0, ..., 1)^T ) (the nth entry is 1, all others are 0). When we compute ( R\mathbf{x} ), the entry corresponding to the all-zero row will be 0 (since every element in that row is 0). This means ( R\mathbf{x} = \mathbf{0} ) but ( \mathbf{x} \neq \mathbf{0} )—which violates the definition of an invertible matrix (invertible matrices only map the zero vector to the zero vector). Thus, ( R ) has no all-zero rows; every row is non-zero.Each pivot must lie on the main diagonal
Since ( R ) is n×n with n non-zero rows, we have exactly n pivots. By the RREF property, pivot columns are strictly increasing. The first pivot has to be in column 1 (there's no column before 1 to place it). The second pivot must be in a column after 1, so column 2 (if we chose column 3, we'd run out of columns by the nth row, as we need n pivots in n columns). Following this pattern, the ith pivot must be in column i for every ( 1 \leq i \leq n ).R is exactly the identity matrix
By RREF rules, every pivot is 1, and all other elements in the pivot's column are 0. Since each pivot is on the main diagonal (row i, column i):- The element at position (i,i) is 1 for all i.
- All off-diagonal elements are 0 (every column's only non-zero element is the diagonal pivot).
This is exactly the structure of the identity matrix ( I ). So we've shown that any invertible RREF matrix must be the identity—no determinants or rank required!
内容的提问来源于stack exchange，提问作者manifold
