If $R$ is invertible, square, and in row-reduced echelon form, then $R = I$. Why?

Question

In their proof of the equivalence of invertibility and row-reduction to an elementary matrix (their Theorem 12 in the 2nd edition, for those who want to reference it), Hoffman and Kunze write

Since $R$ is a (square) row-reduced echelon matrix, $R$ is invertible if and only if each row of $R$ contains a non-zero entry, that is, if and only if $R = I$.

I am struggling with the "only if" in the above. The "if" is immediate, for if each row of $R$ contains a non-zero entry then by the fact that $R$ is square and the definition of row-reduced echelon form, we must have $R = I$, and thus $R$ is plainly invertible ($I$ is its inverse). I can't see how to argue the "only if". I am hoping that an answer will not use any extra machinery, but simply use the basic definitions referenced in the claim above. Essentially, I am looking for a proof of the following:

If $R$ is invertible, square, and in row-reduced echelon form, then $R = I$.

Answer 1

If some row does not contain a non-zero entry, i.e. if some row is zero, then $R$ is clearly not invertible:

if the $k$-th row is zero, then $E_{k,k}R=0,$ where the matrix $E_{k,k}$ has all entries equal to $0$ except a $1$ in position $(k,k).$ Therefore, there is no matrix $S$ such that $RS=I$ because that would entail $$0=0S=E_{k,k}RS=E_{k,k}I=E_{k,k}.$$

Answer 2

Because $R$ is in row echelon form, the $(1,1)$ entry must be either $0$ or $1$, and every other entry in the first column must be $0$. Because $R$ is invertible, the entry must be a $1$ (otherwise you have a column of zeros). So the matrix looks like $$R = \left(\begin{array}{ccccc} 1 & * & * &\cdots & *\\ 0 & * & * &\cdots & *\\ 0 & * & * & \cdots & *\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & * & * & \cdots & * \end{array}\right).$$ Because $R$ is in row echelon form, the $(2,2)$ entry is either $0$ or $1$, and the entries below it must be $0$.

If the entry is $0$, then either the entire row is $0$, or the pivot lies in a column after the second column. That forces the pivot in the third row to be after the third column, and so on. Which means that run out of columns before you run out of rows, so the last row would have to consist of only $0$s. That is impossible, because $R$ is invertible. We conclude that the $(2,2)$ entry is $1$.

And since $R$ is in reduced row echelon form, the entries above the $(2,2)$ entry must also be $0$. So the matrix looks like $$R = \left(\begin{array}{ccccc} 1 & 0 & * &\cdots & *\\ 0 & 1 & * &\cdots & *\\ 0 & 0 & * & \cdots & *\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & * & \cdots & * \end{array}\right).$$ Arguing the same way we obtain that the $(3,3)$ entry is $1$, and the remaining entries in the third column are $0$. Continuing this way, we conclude that $R$ is the identity matrix, as claimed.

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To see that an invertible matrix cannot have a row or columns of zeros, note that the $(i,j)$ entry of $AB$ is $$a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{ik}b_{kj},$$ for suitable $k$. If a row of $A$ is all zeros, then $a_{ir}=0$ for all $r$, so all entries in the $i$th row of $AB$ are zero; thus $AB$ is never the identity matrix. If the $j$th column of $A$ is all zeros, then $a_{sj}=0$ for all $s$, so every entry in the $j$th column of $BA$ is $0$; thus, $BA$ is never the identity matrix. Thus, a matrix with a row of zeros or a column of zeros cannot be invertible.