I'm reading Linear Algebra by Hoffman, Kunze where the authors explained that a $n\times n$ matrix $A$ being invertible is equivalent to the fact that $A$ is row-equivalent to $n\times n$ matrix $R$ which is an identity matrix.

In the proof of the theorem, they wrote:

$$R= E_k\ldots E_2E_1 A$$ where $E_1,\ldots,E_k$ are elementary matrices. Each $E_j$ is invertible, and so $$A = E_1^{-1}\ldots E_k^{-1}~R\,. $$ ... Since, $R$ is a (square) row-reduced echelon matrix, $R$ is invertible if and only if $R=I\,.$ [...]

I couldn't get the conclusion, since any row of $R$ can't be zero, it has to be identity matrix $I\,.$ Why is it so?

Isn't there any other row-reduced echelon matrix other than the identity matrix having no zero row and invertible? Why is it so?


Suppose that $R$ is a matrix in row-reduced echelon form, and that $R$ has no zero-rows. That means that $R$ has a pivot (leading $1$) in every row.

This means that we have $n$ pivots in an $n \times n$ matrix. However, since no column of a row-reduced matrix can have two pivots, it must be that every single column has a pivot. In other words, every column has a leading $1$ in some entry, and the other entries of that column of zero.

In other words, the columns of $R$ must be the columns of the identity. The only order we can put those columns in and have $R$ in row-echelon form is the order in which $R = I$.

  • $\begingroup$ I do not understand your answer. You say that "every column has a leading $1$ in some entry, and the other entries of that column of zero.". But why can't the row reduced echelon form have a zero row? $\endgroup$ – ab123 Sep 3 '18 at 11:08

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