# Why does applying a elementary row / column operation to $I$ give us the corresponding elementary matrix?

Theoretically, why does this work? How do we know the resulting matrix performs the operation on ALL other matrices?

I guess what I'm asking is whether you can prove without "checking" the resulting matrix that it performs the same operation on all other matrices.

So something like : Let $E_o$ be an elementary operation. Then applying $E_o$ to $I$, we get

$I -> E_m$, where $E_m$ is the matrix associated with the elementary matrix.

Three questions :

1) How do we know $I$ functions as the "identity" when we're applying these operations? All we know is that it works that way when we multiply.

2) How do we know $E_m$ performs the operation we applied?

3) How do we know $E_m$ works for all matrices?

It's really crucial that the proof doesn't explicitly find the resulting matrix and then just check that it does what it's supposed to for all matrices.

• The definition of matrix multiplication is such that the $i$th row of $AB$ is a linear combination of the rows of $B$ with coefficients given in the $i$th row of $A$. Furthermore, this multiplication is associative. – Christoph Feb 26 '18 at 12:24


Now I'm going to define a new operation $T$ on matrices: $$T(\mm) = \me \cdot \mm.$$

And your question, now that I've got the definitions out of the way, is

"How do I know that $$S(\mm) = T(\mm)$$ for every matrix $\mm$?"

The first thing to realize is that what I've written above, without the word 'elementary' in "some other elementary operation $S$", is not enough to guarantee this result. It's possible that $S$ might be defined so that $S(\mi) = \me$, where $\me$ is some non-identity matrix, but $S(\mm) = \mi$ for every $\mm \ne \mi$. In that case, the claim that $S(\mm) = T(\mm)$ would be false.

So we're going to need to use the notion of "elementary" somehow. It turns out that the key property that elementary operations share --- the one that makes this proof go through --- is that $$S(\mm \cdot \mk) = S(\mm) \cdot \mk \tag{2}$$ where the $\cdot$ here represents matrix multiplication. Supposing, for the moment, that our elementary operation has this property, it's easy to prove that $S(\mm) = T(\mm)$ as follows. We know that \begin{align} \mi \cdot \mm &= \mm & \text{property of identity} \\ S(\mi \cdot \mm) &= S(\mm) & \text{apply $S$ to both sides} \\ S(\mi) \cdot \mm &= S(\mm) & \text{use equation 2 above} \\ \me \cdot \mm &= S(\mm) & \text{use equation 1 above} \\ \end{align} and that's the result we wanted, because the left hand side is exactly the definition of $T(\mm)$.

Now how do we prove that fundamental claim, equation 2? We need to use elementary-ness to do so. It's going to come down to cases, alas. For instance, if $S$ is an operation that swaps rows $p$ and $q$, I need to write out $S$ explicitly: $$S(\mm)_{ab} = \begin{cases} m_{ab} & a \ne p \text{ and } a \ne q \\ m_{pb} & a = q \\ m_{qb} & a = p \\ \end{cases}$$ That is to say, the $ab$--entry of $S(\mm)$ is the $ab$-entry of $\mm$, unless the row ($a$) is either $p$ or $q$, in which case you have to swap things around.

Now looking at a pair of matrices $\mm$ and $\mk$, we need to know a formula for the $ab$ entry of the product $\mm \cdot \mk$; that's $$(mk)_{ab} = \sum_i m_{ai} k_{ib}.$$ I'm following the usual convention that the entries of $\mm$ are $m_{ij}$ and those of $k$ are $k_{ij}$, and using the two-letter name "$mk$" (in parens) to indicate entries of the product, so that the $i$th row, $j$th column entry of $\mm \cdot \mk$ is denoted $(mk)_{ij}$.

And now we can look at $S(\mm \cdot \mk)_{ab}$ and $(S(\mm) \cdot \mk)_{ab}$ and compare to see whether they're the same. Well \begin{align} S(\mm \cdot \mk)_{ab} &= \begin{cases} (mk)_{ab} & a \ne p \text{ and } a \ne q \\ (mk)_{pb} & a = q \\ (mk)_{qb} & a = p \end{cases} & \text{definition of S} \\ &= \begin{cases} \sum_i m_{ai} k_{ib} & a \ne p \text{ and } a \ne q \\ \sum_i m_{pi} k_{ib} & a = q \\ \sum_i m_{qi} k_{ib} & a = p \end{cases} & \text{definition of matrix multiply} \\ &=\left( \begin{cases} \sum_i m_{ai} & a \ne p \text{ and } a \ne q \\ \sum_i m_{pi} & a = q \\ \sum_i m_{qi} & a = p \end{cases} \right) k_{ib} & \text{extract k_{ib} factor from each!} \\ &=\sum_i \left( \begin{cases} m_{ai} & a \ne p \text{ and } a \ne q \\ m_{pi} & a = q \\ m_{qi} & a = p \end{cases} \right) k_{ib} & \text{extract \sum_i, by distributive law} \\ &= \sum_i S(\mm)_{ai} k_{ib} & \text{replace large parens with S(\mm)_{ai}, because they match} \\ &= (S(\mm) \cdot \mk)_{ab}. \end{align} So they are, in fact, equal, and we're done (with this case).

That still leaves the "multiply a row by a constant" case, and the "add $r$ times one row to another" case to work out, but the proofs for those cases are nearly identical to this one, so I leave that to you.

• Ah, so you do need to explicitly write down what the elementary operation does.. That's a lot of effort you put in when most would have just said something like "We can prove this claim by noting what a elementary operation does case-by-case", so thank you. – Saad Feb 26 '18 at 16:49
• You're welcome. As usual, I learned something by writing this out. There's probably a simple, more abstract, approach, but sometimes getting really concrete has its advantages, too. – John Hughes Feb 26 '18 at 17:16


Here's the alternative approach.

1. Recognize that the set of all $n \times n$ matrices, which I'll call $V$, is itself a vector space with the operations of matrix addition and scalar multiplication.

2. Observe that every elementary row operation $S: V \to V$ is linear: for instance, if you swap two rows of $\mm + \mk$, it's the same as swapping those rows in $\mm$, and swapping them in $\mk$, and then adding the results. In formulas, $$S(\mm + \mk) = S(\mm) + S(\mk).$$ Similarly, if we triple $\mm$ and swap two rows, it's the same as swapping two rows and then tripling the result, i.e., $$S(\alpha\mm) = \alpha S(\mm).$$ So $S$ is linear.

3. Let $\mf_{ij}$ be a matrix that is all zero except the $ij$ element, which is one. Then the set of all $\mf_{ij}$, for $1 \le i,j \le n$, forms a basis for $V$. Any two linear maps whose value on this basis are the same must actually be identical.

4. Now define $T$ as I did above: let $\me = S(\mi)$, and define $T(\mm) = \me \cdot \mm$. For each of the elementary operations, you can write down $\me$ pretty easily: they're matrices that almost look like the identity, except that two rows might be swapped, or one diagonal element might not be $1$, or one off-diagonal element might not be one.

5. For each basis matrix $F_{ij}$, compare $T(F_{ij}) = \me \cdot F_{ij}$, which is pretty easy to compute, with $S(F_{ij})$. Observe that for every $ij$ pair, these two are equal.

6. Hence, by the observation that linear maps agreeing on a basis are equal maps, we have that $T(\mm) = S(\mm)$ for all $\mm$.

Step 5 is the ugly one that actually requires you to do some matrix multiplying (indeed, requires you to do $n^2$ different matrix multiplies!), so you don't really get away from some real work. But by using theorems about linearity, we move away from the purely computational rules and the axioms of the real numbers into the slightly more abstract realm of linear transformations.