## 特殊矩陣 (15)：Pascal 矩陣 (下)

$\displaystyle P(t)=e^{Ht}=\sum_{k=0}^{\infty}\frac{t^k}{k!}H^k$

$\displaystyle P(t)=\sum_{k=0}^{n-1}\frac{t^k}{k!}H^k$

$P(s)P(t)=P(s+t)$

$s=-t$，即得 $P(-t)P(t)=P(0)=I$，故 $P^t$ 是可逆矩陣，且 $(P^t)^{-1}=P(t)^{-1}=P(-t)$；當 $t=1$，就得到帕斯卡逆矩陣 $P^{-1}=P(-1)$。下面是 $n=4$ 的例子：

$H=\begin{bmatrix} 0&0&0&0\\ 1&0&0&0\\ 0&2&0&0\\ 0&0&3&0 \end{bmatrix},~P(t)=\begin{bmatrix} 1&0&0&0\\ t&1&0&0\\ t^2&2t&1&0\\ t^3&3t^2&3t&1 \end{bmatrix},$

$P=\begin{bmatrix} 1&0&0&0\\ 1&1&0&0\\ 1&2&1&0\\ 1&3&3&1 \end{bmatrix},~P^{-1}=\left[\!\!\begin{array}{rrrc} 1&0&0&0\\ -1&1&0&0\\ 1&-2&1&0\\ -1&3&-3&1 \end{array}\!\!\right]$

$P(t)=D(t)PD(t)^{-1}$

$P^{-1}=P(-1)=D(-1)PD(-1)^{-1}$

$\displaystyle \sum_{k=j}^i\binom{i-1}{k-1}\binom{k-1}{j-1}=2^{i-j}\binom{i-1}{j-1},~~~i\ge j$

$\displaystyle \sum_{k=1}^{n+1}\binom{n}{k-1}\binom{k-1}{0}=\sum_{k=0}^n\binom{n}{k}=2^n$

$\displaystyle \sum_{k=0}^nk\binom{n}{k}=n2^{n-1}$

$\displaystyle \sum_{k=0}^n\frac{k(k-1)}{2}\binom{n}{k}=\frac{n(n-1)}{2}2^{n-2}$

\displaystyle\begin{aligned} \sum_{k=0}^nk^2\binom{n}{k}&=(n^2-n)2^{n-2}+\sum_{k=0}^nk\binom{n}{k}\\ &=(n^2-n)2^{n-2}+n2^{n-1}\\ &=(n^2+n)2^{n-2},\end{aligned}

$\displaystyle \sum_{k=j}^i(-1)^k\binom{i-1}{k-1}\binom{k-1}{j-1}=(-1)^j\delta_{ij},~~~i\ge j$

$\displaystyle \sum_{k=0}^{n}(-1)^k\binom{n}{k}=0$

$\left[\!\!\begin{array}{rrrc} 1&0&0&0\\ -1&1&0&0\\ 0&-1&1&0\\ 0&0&-1&1 \end{array}\!\!\right]\begin{bmatrix} 1&0&0&0\\ 1&1&0&0\\ 1&2&1&0\\ 1&3&3&1 \end{bmatrix}=\begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&1&1&0\\ 0&1&2&1 \end{bmatrix}$

\displaystyle\begin{aligned} (EP_n)_{ij}&=(P_n)_{i,j}-(P_{n})_{i-1,j}\\ &=\binom{i-1}{j-1}-\binom{i-2}{j-1}=\binom{i-2}{j-2}\\ &=(P_{n})_{i-1,j-1},\end{aligned}

$EP_n=\begin{bmatrix} 1&\mathbf{0}^T\\ \mathbf{0}&P_{n-1}\end{bmatrix}$

$E^{-1}=\begin{bmatrix} 1&0&0&0\\ 1&1&0&0\\ 1&1&1&0\\ 1&1&1&1 \end{bmatrix}$

$\displaystyle (E^{-1}\hat{P}_{n})_{ij}=\sum_{k=1}^n(E^{-1})_{ik}(\hat{P}_n)_{kj}=\sum_{k=j}^i\binom{k-2}{j-2}=\binom{i-1}{j-1}$

$S=\begin{bmatrix} 1&0&0&0\\ 1&1&0&0\\ 1&2&1&0\\ 1&3&3&1 \end{bmatrix}\begin{bmatrix} 1&1&1&1\\ 0&1&2&3\\ 0&0&1&3\\ 0&0&0&1 \end{bmatrix}=\left[\!\!\begin{array}{ccrr} 1&1&1&1\\ 1&2&3&4\\ 1&3&6&10\\ 1&4&10&20 \end{array}\!\!\right]$

$\displaystyle \sum_{k=1}^np_{ik}p_{jk}=\sum_{k=1}^n\binom{i-1}{k-1}\binom{j-1}{k-1}=\binom{i+j-2}{j-1}=s_{ij}$

$s_{ij}=s_{i-1,j}+s_{i,j-1}$

$EP_nP_n^TE^T=(EP_n)(EP_n)^T=\begin{bmatrix} 1&\mathbf{0}^T\\ \mathbf{0}&P_{n-1} \end{bmatrix}\begin{bmatrix} 1&\mathbf{0}^T\\ \mathbf{0}&P_{n-1}^T \end{bmatrix}=\begin{bmatrix} 1&\mathbf{0}^T\\ \mathbf{0}&S_{n-1} \end{bmatrix}$

$(ES_n)_{ij}=(S_n)_{ij}-(S_n)_{i-1,j}$

$(ES_nE^T)_{ij}=((S_n)_{ij}-(S_n)_{i-1,j})-((S_n)_{i,j-1}-(S_n)_{i-1,j-1})=(S_n)_{i-1,j-1}$

$\displaystyle \sum_{k=1}^n\binom{i-1}{k-1}^2=\sum_{k=1}^{i-1}\binom{i-1}{k-1}^2=\binom{2i-2}{i-1}=i\cdot c_{i-1}$

\begin{aligned} S^{-1}&=(PP^T)^{-1}=(P^T)^{-1}P^{-1}\\ &=(DP^TD^{-1})(DPD^{-1})=DP^TPD^{-1}\\ &=(DP^T)(PP^T)(DP^T)^{-1}=(DP^T)S(DP^T)^{-1},\end{aligned}

There are so many relations present that when someone finds a new identity, there aren’t many people who get excited about it any more, except the discoverer!

[1] L. Aceto and D. Trigiante, The Matrices of Pascal and Other Greats, The American Mathematical Monthly, Vol. 108, 2001, pp 232-245.
[2] A. Edelman and G. Strang, Pascal matrices, The American Mathematical Monthly,Vol. 111, 2004, pp 189-197.

### 3 Responses to 特殊矩陣 (15)：Pascal 矩陣 (下)

1. 延伸寸 說道：

Sing for the sake of singing.

• ccjou 說道：

我想起電影「阿甘正傳」裡 Forrest Gump 講過的一段話(解釋他為何不停地跑步)：

That day, for no particular reason, I decided to go for a little run. So I ran to the end of the road. And when I got there, I thought maybe I’d run to the end of town. And when I got there, I thought maybe I’d just run across Greenbow County. And I figured, since I run this far, maybe I’d just run across the great state of Alabama. And that’s what I did. I ran clear across Alabama. For no particular reason I just kept on going. I ran clear to the ocean. And when I got there, I figured, since I’d gone this far, I might as well turn around, just keep on going. When I got to another ocean, I figured, since I’d gone this far, I might as well just turn back, keep right on going.

2. ccjou 說道：

YouTube 有這段 i just felt like running 影片：