Exercise 2.1.1

Since we have $S_{i,i-1}=1$ for $i=2,\dots ,n$, so for matrix $D$, we have $d_{i,i}=1$ for $i=1,\dots ,n$ and $d_{i,i-1}=-1$ for $i=2,\dots ,n$.

With column times rows we have:

$S{S}^T={\sum <!--\; FUNCTION\; APPLICATION\; -->}_i{s}_i{s}_i^T$ where for $\mathit{in}$, $s_i=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill ⋮\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \\ \hfill ⋮\hfill \\ \hfill 0\hfill \end{array}\right]$, the 1 appears at the $i+1$ position and $s_n=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill ⋮\hfill \\ \hfill 0\hfill \end{array}\right]$.

So for $\mathit{in}$, we have $K^i=s_i{s}^T{s}_i$, so $K_{i+1,i+1}^i=1$ while all other entries are zero. And for $i=n$, we have $K^n=0$. So combine all $K^i$ together, we have $S{S}^T=I-K^0=I-\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill \dots \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \dots \hfill \\ \hfill \dots \hfill & \hfill \dots \hfill & \hfill \dots \hfill \\ \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \end{array}\right]$

So we have $D{D}^T=(I-S){(I-S)}^T=(I-S)(I-S^T)=I-S^T-S+S{S}^T=-S+2I-S^T-K^0=A-K^0$.

So $D{D}^T$ equals $A$ except that the former has 1 and the latter has 2 in their $(1,1)$ entries.

Similarly we can see that $D^{T}D$ equals to $A$ except that $1\ne 2$ in their $(n,n)$ entries.