My question is very simple - Can an elementary row operation change the size (eg: $2\times2$ or $3\times 2$) of a matrix?

I think the answer should be no, but while reading Linear Algebra by Hoffman Kunze I stumbled upon this:

Definition. An $m\times n$ matrix is said to be an elementary matrix if it can be obtained from the $m\times m$ identity matrix by means of a single elementary row operation.

Now, I know of 3 elementary row operations, (adding a multiple of one row to another, multiplying throughout a row by a non zero constant and interchanging two rows) but none of them can change the size of a matrix.

But since this is a highly praised book I can't trust myself as much as I'd like to.

  • 6
    $\begingroup$ Typo in the definition. It should be $n=m$. $\endgroup$ May 26, 2016 at 17:39
  • $\begingroup$ Are there on-line errata for Hoffman & Kunze? or is it too ancient a book for that. $\endgroup$
    – GEdgar
    May 27, 2016 at 15:13
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    $\begingroup$ @Dietrich I disagree that it's necessarily a typo. It could just be an unnecessarily general definition from which it would follow as an easy theorem that elementary matrices are square (since row operations don't change dimensions). $\endgroup$
    – Mark S.
    May 27, 2016 at 15:16
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    $\begingroup$ In any case the elementary matrices are square matrices, either as a consequence of this definition or by dint of a simpler definition. We intend left multiplication by an elementary matrix to accomplish the result of the elementary row operation. $\endgroup$
    – hardmath
    May 27, 2016 at 16:38
  • $\begingroup$ @GEdgar Nothing official I think but there's this site post: math.stackexchange.com/questions/437253/… $\endgroup$
    – Aritra Das
    May 27, 2016 at 18:33

1 Answer 1


Original Answer:

It must be a typo. For another reference, if you look at Horn & Johnson's book here (chapter 0, section 0.3.3 in the first edition) the authors discuss how elementary row operations can be achieved via left multiplication by square matrices.

Side note: If we use the fact that elementary row operations on a matrix $\boldsymbol{M}$ are equivalent to multiplying $\boldsymbol{M}$ on the left by (certain) square matrices, it is easy to determine the effect of elementary row operations on the determinant (recall that, for square matrices, $det(\boldsymbol{UM})=det(\boldsymbol{U})det(\boldsymbol{M})$).


I can see it not being a typo if what the authors mean is this:

Elementary row operations can be represented using matrix multiplication. How? Suppose I have a matrix $\boldsymbol{A}\in M_{m\times n}$ and I want to apply any of the three operations. All I have to do is:

  • Begin with the identity matrix of size $\boldsymbol{m}$, call it $\boldsymbol{I}_m$.
  • Apply whichever of the three elementary row operations you want to $\boldsymbol{I}_m$, call this new matrix $\boldsymbol{I}_{new}$. In other words, if you want to apply an elementary row operation to $\boldsymbol{A}$ you first apply it to $\boldsymbol{I}_m$.
  • Multiply $\boldsymbol{A}$ on the left by $\boldsymbol{I}_{new}$.

Note that this method for performing elementary row operations essentially begins with the identity matrix $\boldsymbol{I}_m$ and then performs the elementary operation on $\boldsymbol{I}_m$. This might be in line with what the authors meant. Of course, you then need to multiply $\boldsymbol{A}$ by this modified identity matrix which won't change the dimension of $\boldsymbol{A}$.

Example (interchanging to rows): I want to change the first and third rows of $\boldsymbol{A}\in \mathbb{R}_{3\times5}$.

Solution: Multiply: \begin{align*} {\underbrace{\left[\begin{array}{ccc}0 & 0 & 1 \\0 & 1 & 0 \\1 & 0 & 0\end{array}\right]}_{\boldsymbol{I}_{new}}}\boldsymbol{A}, \end{align*} this will change the first and third row of $\boldsymbol{A}$.

Note: the determinant of $\boldsymbol{I}_{new}$ here is -1. (This can be shown using the Alternating Sums formulation of the determinant).

Example (multiplying a row by a non-zero scalar): I want to multiply the second row of $\boldsymbol{A}\in \mathbb{C}_{3\times 5}$ by 7/3.

Solution: Multiply: \begin{align*} {\underbrace{\left[\begin{array}{ccc}1 & 0 & 0 \\0 & 7/3 & 0 \\0 & 0 & 1\end{array}\right]}_{\boldsymbol{I}_{new}}}\boldsymbol{A}, \end{align*} this will multiply the second row of $\boldsymbol{A}$ by $7/3$.

Note: The determinant of $\boldsymbol{I}_{new}$ here is 7/3. (The determinant of a diagonal matrix is the product of the diagonal entries.)

Example (adding a multiple of one row to another): I want to add 5 times row 1 to row 2 of $\boldsymbol{A}\in \mathbb{C}_{3\times 5}$.

Solution: Multiply: \begin{align*} {\underbrace{\left[\begin{array}{ccc}1 & 0 & 0 \\5 & 1 & 0 \\0 & 0 & 1\end{array}\right]}_{\boldsymbol{I}_{new}}}\boldsymbol{A}, \end{align*} this will multiply the first row of $\boldsymbol{A}$ by $5$ and add it to the second row of $\boldsymbol{A}$ leaving the first row unchanged.

Note: The determinant of $\boldsymbol{I}_{new}$ here is 1. (The determinant of a triangular matrix is the product of the diagonal entries.)

Notice that, in each of these cases, all that was necessary was to choose the identity matrix of the right dimension (in our case, $3\times3$) and to apply the elementary row operation to this identity matrix. Then, we multiply $\boldsymbol{A}$ on the left by this new matrix.

  • If we wish to apply multiple elementary row operations then we can simply repeat this process sequentially for each operation we apply.

  • It is now easy to see how elementary row operations change the determinant of a square matrix $\boldsymbol{A}$ since the determinant of the product of two matrices is the product of the determinants. Additionally, the determinants of the "$\boldsymbol{I}_{new}$" matrices are easy to compute.

  • All these "$\boldsymbol{I}_{new}$" matrices are nonsingular (it is easy to see that their determinant is nonzero).


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