## Matrix Multplication

December 2020

Theorem. Let $$e$$ be an elementary row operation and let $$E$$ be the $$m \times m$$ elementary matrix $$E = e(I)$$. Then, for every $$m \times n$$ matrix $$A$$,

$e(A) = EA$

Proof. We prove this for each of the three elementary row operations separately.

Swap: If $$e$$ 'swaps' rows $$s$$ and $$r$$ of $$A$$, then

$e(A)_{ij} = \begin{cases} A_{ij} \quad \text{if } s \neq i \neq r \\ A_{sj} \quad \text{if } i = r \\ A_{rj} \quad \text{if } i = s \end{cases}$

and

$EA_{ij} = e(I)A_{ij} = \sum\limits_{k=1}^{m} e(I)_{ik}A_{kj}.$

Noting that the same row operation as before is being applied to $$I$$, we can equivalently define this matrix as follows:

$e(I)A_{ij} = \begin{cases} \sum\limits_{k=1}^{m} I_{ik}A_{kj} \quad \text{if } s \neq i \neq r \\ \sum\limits_{k=1}^{m} I_{sk}A_{kj} \quad \text{if } i = r \\ \sum\limits_{k=1}^{m} I_{rk}A_{kj} \quad \text{if } i = s \end{cases}$

We can now simplify each of these cases. For the first case, since $$I_{ik} = 0$$ for all but the $$i$$-th column, the only non-zero term in the sum will be $$I_{ii}A_{ij} = A_{ij}$$. For the second case, the only non-zero term in the sum will be $$I_{ss}A_{sj} = A_{sj}$$ and for the third case the only non-zero term in the same will be $$I_{rr}A_{rj} = A_{rj}$$.

Scale: If $$e$$ scales some row $$r$$ by some scalar $$c$$, then

$e(A)_{ij} = \begin{cases} A_{ij} \quad \text{if } i \neq r \\ cA_{rj} \quad \text{if } i = r \end{cases}$

and

$EA_{ij} = e(I)A_{ij} = \sum\limits_{k=1}^{m} e(I)_{ik}A_{kj}.$

Noting that the same row operation as before is being applied to $$I$$, we can equivalently define this matrix as follows:

$e(I)A_{ij} = \begin{cases} \sum\limits_{k=1}^{m} I_{ik}A_{kj} \quad \text{if } s \neq i \neq r \\ \sum\limits_{k=1}^{m} c I_{rk}A_{kj} \quad \text{if } i = r \end{cases}$

We've already seen why the first case simplifies to $$A_{ij}$$, and for identical reasons the second case simplifies to $$c A_{rj}$$.

Add a scaled row to another row: If $$e$$ scales adds row $$s$$ of the matrix, scaled by a scalar $$c$$, to another row $$r$$ (where $$r \neq s$$), then

$e(A)_{ij} = \begin{cases} A_{ij} \quad \text{if } i \neq r \\ A_{rj} + cA_{sj} \quad \text{if } i = r \end{cases}$

and

$EA_{ij} = e(I)A_{ij} = \sum\limits_{k=1}^{m} e(I)_{ik}A_{kj}.$

If $$i \neq r$$, then $$e(I)A_{ij} = A_{ij}$$, as we have shown before. If $$i = r$$, then the only non-zero terms of the sum above are $$e(I)_{rr}A_{rj}$$ and $$e(I)_{rs}A_{sj}$$. Since $$I_{ss} = 1$$ and $$e(I)_{rs} = cI_{ss}$$, $$e(I)_{rs} = c$$. Then the value of the sum above if $$i=r$$ is $$A_{rj} + cA_{sj}$$.