1

Notation has always been my downfall when it comes to writing proofs. I have been given a general problem to try and write a proof based on this. This is just for research and practice. Given a matrix $e_{ij}A_{nxm}$ If the matrix is multiplied on the left by

$$\begin{bmatrix}0&1\\0&0\end{bmatrix}$$ multiplied by $$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ the result is $$\begin{bmatrix}c&d\\0&0\end{bmatrix}$$

Which tells me that row j replaces row i and everything else becomes zeroes

Multiplication on the left by $$\begin{bmatrix}0&0\\1&0\end{bmatrix}$$ times $$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ produces $$\begin{bmatrix}0&0\\a&b\end{bmatrix}$$

which tells me that this kind of multiplication will put row i to row j and everything else will become zeroes. I'm just not sure how to officially write this as a proof.

cele
  • 2,247
  • Try to think in terms of linear combination, notice that every row of your product is the linear combination of the first row from $e_{ij}$ with each entry of each column $j$. – Aaron Maroja Sep 08 '14 at 23:32
  • For that second product, you should have $\pmatrix{0&0\a&b}$ – Ben Grossmann Sep 08 '14 at 23:45

1 Answers1

1

Remember the basic product formula: if $A=[a_{hk}]$ is an $m\times n$ matrix and $B=[b_{hk}]$ is an $n\times p$ matrix, then the coefficient in place $(h,k)$ of $AB$ is $$ \sum_{l=1}^n a_{hl}b_{lk}. $$

The $n\times n$ matrix $E_{ij}$ having $1$ in place $(i,j)$ can be described by having coefficients $c_{hk}$ $(1\le h\le n$, $1\le k\le n)$ where $$ c_{hk}=\delta_{hi}\delta_{jk} $$ where, as usual, $$ \delta_{rs}=\begin{cases} 1&\text{if $r=s$,}\\ 0&\text{if $r\ne s$.} \end{cases} $$ Consider an $n\times n$ matrix $A=[a_{hk}]$; the coefficient $b_{hk}$ in place $(h,k)$ of the product $B=E_{ij}A$ can be written as \begin{align} b_{hk}=\sum_{l=1}^n (\delta_{hi}\delta_{lj})a_{lk} &=\sum_{l=1}^n \delta_{hi}(\delta_{lj}a_{lk})\\ &=\sum_{l=1}^n \delta_{hi}(\delta_{lj}a_{lk})\\ &=\delta_{hi}\sum_{l=1}^n (\delta_{lj}a_{lk})\\ &=\delta_{hi}a_{jk} \end{align} Therefore $$ b_{hk}=\begin{cases} 0&\text{if $h\ne i$,}\\ a_{jk}&\text{if $h=i$.} \end{cases} $$ This amounts exactly to saying that all coefficients of $B$ are zero except possibly for row $i$, which is the same as row $j$ of $A$.

egreg
  • 238,574