Full-rank square matrix in RREF is the identity matrix
Dependencies:
Let $A$ be an $n$ by $n$ matrix in RREF. Then all rows of $A$ are non-zero iff $A = I_n$ (identity matrix).
Proof
If $A$ is the identity matrix, all rows of $A$ are non-zero.
Let all rows of $A$ be non-zero. Let $\alpha_i$ be the smallest value of $j$ such that $A[i, j] \neq 0$. Since $A$ is in RREF, $\alpha_i < \alpha_{i+1}$. Therefore, $1 \le \alpha_1 < \alpha_2 < \ldots < \alpha_n \le n$. Therefore, $\alpha_i = i$.
Since $A$ is in RREF, $A[i, i] = A[i, \alpha_i] = 1$ and there is only one non-zero entry in the $\alpha_i^{\textrm{th}}$ column.
Therefore, $A$ is the identity matrix.
Dependency for:
- RREF([A|I]) = [I|inv(A)] iff A is invertible
- Full-rank square matrix is invertible
- A matrix is full-rank iff its determinant is non-0
- Determinant of product is product of determinants
Info:
- Depth: 4
- Number of transitive dependencies: 6