This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
Distributive law and Laplace expansion
This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
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from lingeo import random_int_list
The determinant function satisfies the distributive law on each row.
That is,
Since one may swap two rows or take the transpose without chaning the determinant, the distributive law holds for any row and any column.
If we think of the determinant $\det(A)$ as a function on $n$ row vectors $f(\br_1,\ldots,\br_n)$,
then $f$ is linear on each vector individually.
That is,
A function with this property is said to be multilinear.
Let $A = \begin{bmatrix} a_{ij} \end{bmatrix}$ be an $n\times n$ matrix.
Let $\br_1, \ldots, \br_n$ be the rows of $A$.
Define $A(i,j)$ as the submatrix obtained from $A$ by removing the $i$-th row and the $j$-th column.
By expanding the first row, we have
$$ \begin{aligned} \det(A) &= \det\begin{bmatrix} a_{11} & \cdots & a_{1n} \\ - & \br_2\trans & - \\ ~ & \vdots & ~ \\ - & \br_n\trans & - \end{bmatrix} \\ &= \det\begin{bmatrix} a_{11} & 0 & \cdots & ~ & 0 \\ - & ~ & \br_2\trans & ~ & - \\ ~ & ~ & \vdots & ~ & ~ \\ - & ~ & \br_n\trans & ~ & - \end{bmatrix} + \det\begin{bmatrix} 0 & a_{12} & 0 & \cdots & 0 \\ - & ~ & \br_2\trans & ~ & - \\ ~ & ~ & \vdots & ~ & ~ \\ - & ~ & \br_n\trans & ~ & - \end{bmatrix} + \det\begin{bmatrix} 0 & \cdots & ~ & 0 & a_{1n} \\ - & ~ & \br_2\trans & ~ & - \\ ~ & ~ & \vdots & ~ & ~ \\ - & ~ & \br_n\trans & ~ & - \end{bmatrix} \\ &= a_{11}\det A(1,1) - a_{12}\det A(1,2) + \cdots + (-1)^{1+n}a_{1n}\det A(1,n). \end{aligned} $$Again, this formula works for any row and column.
We may expand the $i$-th row and get
The formula for expanding a column is similar.
This identity is called the Laplace expansion.
### code
set_random_seed(0)
print_ans = False
n = 4
A = matrix(n, random_int_list(n^2,3))
expr = []
dets = []
for j in range(n):
expr.append(LatexExpr(r"\det"))
B = copy(A)
for k in range(n):
if k != j:
B[0,k] = 0
dets.append(B.det())
expr.append(copy(B))
if j != n-1:
expr.append(LatexExpr("+"))
print("n =", n)
pretty_print(LatexExpr(r"\det"), A, LatexExpr("="), *expr)
if print_ans:
print("%s = "%A.det() + " + ".join("%s"%d for d in dets))
計算題目給的等式中的每一個行列式值、
並驗證等式成立。
Calculate each term in the equality and verify the equality holds.
以下題目探討矩陣中單一一項對行列式值的影響。
The following problem studies how an entry of a matrix changees the determinant.
令
$$ A = \begin{bmatrix} 1 + x & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}. $$求 $\det(A)$。
Let
$$ A = \begin{bmatrix} 1 + x & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}. $$Find $\det(A)$.
觀察到
$$ (1+x,2,3) = (1,2,3) + (x,0,0). $$利用這個性質，
將 $\det(A)$ 的常數項寫成一個 $3\times 3$ 矩陣的行列式值、
並將其一次項寫成一個 $2\times 2$ 矩陣的行列式值。
Use the fact that
$$ (1+x,2,3) = (1,2,3) + (x,0,0) $$to write $\det(A)$ as a polynomial of degree $2$ such that its constant term is the determinant of a $3\times 3$ matrix while its linear term is the determinant of a $2\times 2$ matrix.
把 $\det(A)$ 看成 $x$ 的函數，求 $\frac{d\det(A)}{dx}$。
搭配上一題，
這說明當一個矩陣 $A$ 的 $i,j$-項增加 $x$ 的時候，
其行列式值會增加 $x\cdot (-1)^{i+j}\det A(i,j)$。
Consider $\det(A)$ as a function of $x$. Find $\frac{d\det(A)}{dx}$.
Use the result of the previous problem, explain the determinant of $A$ increases by $x\cdot (-1)^{i+j}\det A(i,j)$ when its $i,j$-entry increases by $x$.
對以下 $n\times n$ 矩陣，
將每列寫成兩個向量相加，
其中一個向量只有常數，另一個向量只有 $x$。
如此可以將 $\det(A)$ 寫成 $2^n$ 個行列式值相加。
計算這些行列式值並計算 $\det(A)$。
For each of the following $n\times n$ matrix, write each row as the sum of two vectors, where one is composed of constants only, while the other is composed of linear terms.
Thus, we may write $\det(A)$ as the sum of $2^n$ determinants. Calculate each of them to find $\det(A)$.
令
$$
A = \begin{bmatrix}
a & b & c & d \\
e & f & g & h \\
i & j & k & \ell \\
m & n & o & p
\end{bmatrix}.
$$
求 $\det(A)$。
給定向量 $\ba$、$\bb$、以及 $\br_2,\ldots,\br_n$。
令
依照以下步驟證明行列式值的分配律 $\det(M) = \det(A) + \det(B)$。
Let $\ba$, $\bb$, and $\br_2, \ldots, \br_n$ be some vectors.
Let
Follow the steps below to show the distributive law of the determinant, that is, $\det(M) = \det(A) + \det(B)$.
證明當 $\{\br_2,\ldots,\br_n\}$ 線性相依的時候，$\det(M) = \det(A) = \det(B) = 0$。
因此等式成立。
Show that $\det(M) = \det(A) = \det(B) = 0$ when $\{\br_2,\ldots,\br_n\}$ is linearly dependent, so the equality holds.
假設 $\{\br_2,\ldots,\br_n\}$ 線性獨立。
找一個向量 $\br_1$ 使得 $\beta = \{\br_1,\br_2,\ldots,\br_n\}$ 是 $\mathbb{R}^n$ 的一組基底。
因此可以將 $\ba$ 和 $\bb$ 寫成 $\beta$ 的線性組合：
藉由列運算說明行列式值的分配律成立。
Suppose $\{\br_2,\ldots,\br_n\}$ is linearly independent. Find a vector $\br_1$ such that $\beta = \{\br_1,\br_2,\ldots,\br_n\}$ is a basis of $\mathbb{R}^n$. Thus, both $\ba$ and $\bb$ can be written as linear combinations of $\beta$:
$$ \begin{aligned} \ba &= a_1\br_1 + \cdots + a_n\br_n \\ \bb &= b_1\br_1 + \cdots + b_n\br_n \end{aligned} $$Use row operations to show the distributive law in this case.