Elementary matrix acts on columns
This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
$\newcommand{\trans}{^\top} \newcommand{\adj}{^{\rm adj}} \newcommand{\cof}{^{\rm cof}} \newcommand{\inp}[2]{\left\langle#1,#2\right\rangle} \newcommand{\dunion}{\mathbin{\dot\cup}} \newcommand{\bzero}{\mathbf{0}} \newcommand{\bone}{\mathbf{1}} \newcommand{\ba}{\mathbf{a}} \newcommand{\bb}{\mathbf{b}} \newcommand{\bc}{\mathbf{c}} \newcommand{\bd}{\mathbf{d}} \newcommand{\be}{\mathbf{e}} \newcommand{\bh}{\mathbf{h}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bq}{\mathbf{q}} \newcommand{\br}{\mathbf{r}} \newcommand{\bx}{\mathbf{x}} \newcommand{\by}{\mathbf{y}} \newcommand{\bz}{\mathbf{z}} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\bw}{\mathbf{w}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\nul}{\operatorname{null}} \newcommand{\rank}{\operatorname{rank}} %\newcommand{\ker}{\operatorname{ker}} \newcommand{\range}{\operatorname{range}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Row}{\operatorname{Row}} \newcommand{\spec}{\operatorname{spec}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\idmap}{\operatorname{id}} \newcommand{\am}{\operatorname{am}} \newcommand{\gm}{\operatorname{gm}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\iner}{\operatorname{iner}}$
from lingeo import random_int_list, row_operation_process
Each $n\times n$ matrix can be viewed as a list of column vectors $\{\bu_1,\ldots, \bu_n\}$.
We define the column parallelotope of $A$ as the polytope
spanned by the columns $\{\bu_1,\ldots, \bu_n\}$ of $A$.
Let $\Vol_C(A)$ be the signed volume of the column parallelotope.
Then $\Vol_C(A)$ can be defined through the following rules.
Thus, $\Vol_C(EA) = -\Vol_C(A)$ and we define $\Vol_C(E) = -1$.
Thus, $\Vol_C(EA) = k\Vol_C(A)$ and we define $\Vol_C(E) = k$.
(Note that this statement still holds even when $k = 0$.)
Thus, $\Vol_C(EA) = \Vol_C(A)$ and we define $\Vol_C(E) = 1$.
As a consequence, if a matrix $A$ is invertible and
can be written as the product a sequence of elementary matrices $F_1\cdots F_k$,
then $\Vol_C(A) = \Vol_C(F_1)\cdots\Vol_C(F_k)\Vol_C(I_n) = \Vol_C(F_1)\cdots\Vol_C(F_k)$.
In contrast, if $A$ is not invertible,
then column parallelotope collapses and is flat $\Vol_C(A) = 0$.
From this point of view,
the value of $\Vol_C(A)$ can be both
the signed volume and
the scaling factor changing $\Vol_C(B)$ to $\Vol_C(AB)$ for any $B$.
執行以下程式碼。
已知 $A = F_1\cdots F_k$ 是一群單位矩陣的乘積。
Run the code below. Suppose we know $A = F_1\cdots F_k$ is the product of some elementary matrices.
### code
set_random_seed(0)
print_ans = False
n = 2
while True:
A = matrix(n, random_int_list(n^2, 3))
if A.det() != 0:
break
elems = row_operation_process(A, inv=True)
pretty_print(A, LatexExpr("="), *elems)
R = identity_matrix(n)
k = 1
vols = [1]
for elem in elems[::-1]:
R = elem * R
vols.append(R.det())
vecs = R.columns()
P = polytopes.parallelotope(vecs).plot(
xmin=-3,
xmax=3,
ymin=-3,
ymax=3,
wireframe="black",
fill="lightgreen"
)
for v in vecs:
P += text("%s"%v, v + vector([0,0.5]), fontweight=1000).plot()
show(P, title="$V_{%s}$"%k)
k += 1
if print_ans:
for k,vol in enumerate(vols):
print("The signed volumn (area) of V%s is"%k, vol)
令 $V_0$ 為單位矩陣的行超平行體。
說明如何從 $V_0$ 得到 $V_1$ 並求出 $V_1$ 的面積。
Let $V_0$ be the column parallelotope for the identity matrix. Explain how to obtain $V_1$ from $V_0$ and find the volume of $V_1$.
依序說明如何從 $V_i$ 得到 $V_{i+1}$,
並求出 $V_2,\ldots, V_k$ 的面積。
Explain how to obtain $V_{i+1}$ from $V_i$ and find the volume of $V_{i+1}$ for each $i$. Then find the volume for each of $V_2,\ldots, V_k$.
對以下矩陣 $A$,
用文字描述它的行平行體,並用求出它的有向體積。
For each of the following matrices $A$, use a few sentences to decribe its column parallelotope. Then find its signed volumn.
在 $\Vol_C$ 的定義中,假設我們已經接受了
說明為什麼
中的 $-1$ 是合理的。
In the definition of $\Vol_C$, suppose we already accept the following rules.
Explain the $-1$ in the rule below is reasonable.
利用 $\Vol_C$ 定義中的四條準則,說明以下性質。
Use the four rules in the definition of $\Vol_C$ to expalin the following properties.
若 $A$ 中有兩列向量相等,說明 $\Vol_C(A) = 0$。
If $A$ has two rows in common, then $\Vol_C(A) = 0$.
若 $A$ 中的列向量集合線性相依,說明 $\Vol_C(A) = 0$。
If some rows of $A$ form a linearly dependent set, then $\Vol_C(A) = 0$.
令
$$ A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}}\\ \end{bmatrix}. $$Let
$$ A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}}\\ \end{bmatrix}. $$已知 $A$ 的行向量彼此互相垂直
且長度皆為 $1$。
以直觀的方式猜測 $\Vol_C(A)$。
真的計算看看你的猜測是否正確。
We know the columns of $A$ are mutually orthogonal and of length $1$. Guess what is $\Vol_C(A)$ by intuition. Then check if your intuition is correct by definition.
將 $A$ 矩陣的
第一個行向量伸縮為原來的 $3$ 倍,
第二個行向量伸縮為原來的 $4$ 倍,
第三個行向量伸縮為原來的 $5$ 倍,
並得到 $B$ 矩陣。
以直觀的方式猜測 $\Vol_C(B)$。
真的計算看看你的猜測是否正確。
Construct a new matrix $B$ from $A$ by rescaling the first column by $3$, the second column by $4$, and the third column by $5$. Guess what is $\Vol_C(A)$ by intuition. Then check if your intuition is correct by definition.
積分的概念在於
把許多小方塊(或是其它幾何形狀)的體積累積起來。
當這些小方塊的體積會隨著座標變動的時候則須要乘上一些調整量來確保體積正確。
我們知道笛卡爾座標和極座標轉換時有以下公式
當 $x = r\cos\theta$ 且 $y = r\sin\theta$ 時,
計算
的 $\Vol_C(J)$。
而這個值告訴我們極座標上的一個小方塊
和笛卡爾座標上的一個小方塊
它們的體積的比率為何。
The spirit of integration is to sum up the volumns of small cubes (or other geometric objects). These cubes will change their volumes by a scalar when some transformation is applied. We knew the formula
$$ \int f\,dxdy = \int f rdrd\theta $$for converting a cartesian coordinate system to a polar coordinate system. Let $x = r\cos\theta$ 且 $y = r\sin\theta$. Calculate $\Vol_C(J)$, where
$$ J = \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta} \\ \end{bmatrix}. $$This told us that the ratio between the volumn of a small cube on a polar coordinate system to that on a cartesian coordinate system.