Worksheet 2
Aside from the zero matrix $0$, all of whose entries are $0$, the next simplest matrix is one where exactly one entry is not zero. We let $D(i,j)_m$ denote the $m \times m$ matrix whose only non-zero entry is a $1$ in position $(i,j)$.
Write down all possible $2 \times 2$ matrices $D(i,j)$. How many are there? Can you figure out a general formula for the number of $m \times m$ $D(i,j)$’s?
Any matrix $A$ is a linear combination of $D(i,j)$’s. Indeed, convince yourself that \(A = \sum_{i,j} a_{i,j} D(i,j).\) In this way, if we can understand the behavior of the $D(i,j)$, then we can hope to extend that understanding to a general matrix.
Along this vein, what is a formula for the product \(D(i,j) D(r,s) \ ?\) (Hint: write down the $2 \times 2$ and see if you can spot a pattern in $j$ and $r$ which tells you whether the product is $0$ or not.)
Let’s walk through a proof that, for $i \neq j$, \(L(i,j,c) L(i,j,-c) = I.\)
- First, express $L(i,j,c)$ in terms of $I$, $D(i,j)$, and $c$.
- Now, express $L(i,j,-c)$ in terms of $I$, $D(i,j)$, and $-c$.
- Multiply these expressions together and use matrix algebra to expand the product.
- Use what we learned in 2 to simplify
(You can understand this matrix formula by reasoning through the corresponding row operations. If I add $c$ copies of $\mathbf{R}_i$ to $\mathbf{R}_i$ and then I subtract $c$ copies of $\mathbf{R}_i$, I end up having done nothing. )