# If a Matrix $A$ is Full Rank, then $\rref(A)$ is the Identity Matrix ## Problem 645

Prove that if $A$ is an $n \times n$ matrix with rank $n$, then $\rref(A)$ is the identity matrix.

Here $\rref(A)$ is the matrix in reduced row echelon form that is row equivalent to the matrix $A$. Add to solve later

## Proof.

Because $A$ has rank $n$, we know that the $n \times n$ matrix $\rref(A)$ has $n$ non-zero rows.
This means that all $n$ rows must have a leading 1.

Each leading 1 must be in a distinct column, so we must have that each of the $n$ columns has a leading 1.
In a row echelon matrix, these leading 1s must be arranged to lie on the diagonal.

In a reduced row echelon matrix, each column with a leading 1 has 0s above and below that 1.

These restrictions mean that $\rref(A)$ must be the identity matrix. Add to solve later

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