反矩陣

Matrix inverse

This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.

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from lingeo import random_good_matrix

Main idea

Matrix-matrix multipliciation (by column)

Let $A$ be an $m\times n$ matrix.
Let $B$ be an $n\times \ell$ matrix whose columns are $\bu_1,\ldots,\bu_\ell$.
Then the columns of $AB$ are $A\bu_1, \ldots, A\bu_\ell$.

Recall that an $n\times n$ matrix $A$ is invertible if there is a matrix $B$ such that $AB = BA = I_n$.
Indeed, an $n\times n$ matrix is invertible if and only if it is nonsingular.

invertible $\implies$ nonsingular
Suppose $A$ and $B$ are $n\times n$ matrices such that $AB = I_n$.
Then both $A$ and $B$ are nonsingular.

nonsingular $\implies$ invertible
Suppose $A$ is an $n\times n$ nonsingular matrix.
Let $\be_1,\ldots,\be_n$ be the columns of $I_n$.
Since $A$ is nonsingular and $\Col(A) = \mathbb{R}^n$, the equation $A\bx = \be_i$ has a solution $\bx = \bb_i$ for each $i = 1,\ldots, n$.
Let $B$ be the matrix whose columns are $\bb_1,\ldots,\bb_n$.
Then $AB = I_n$.

Here is a way to calculate $B$ at once:

1. Consider the $n\times 2n$ augmented matrix $\left[\begin{array}{c|c} A & I_n \end{array}\right]$.
2. Since $A$ is nonsingular, necessarily the reduced echelon form of $\left[\begin{array}{c|c} A & I_n \end{array}\right]$ is $\left[\begin{array}{c|c} I_n & B \end{array}\right]$ for some $B$.
3. Thus, $AB = I_n$.

Suppose $A$ and $B$ be $n\times n$ matrices such that $AB = I_n$.
Then $BA = I_n$.
Therefore, if $AB = I_n$, then both $A$ and $B$ are invertible and they are the inverse of each other.

Suppose $A$, $B$, and $C$ are $n\times n$ matrices such that $CA = I_n$ and $AB = I_n$.
Then $C = B$.
Therefore, each matrix only has one inverse.

Side stories

• is_invertible
• inverse
• inverse algebra

Experiments

Exercise 1

執行下方程式碼。
矩陣 $\left[\begin{array}{c|c} I & B \end{array}\right]$ 是 $\left[\begin{array}{c|c} A & I \end{array}\right]$ 的最簡階梯形式矩陣。

Run the code below. Let $\left[\begin{array}{c|c} I & B \end{array}\right]$ be the reduced echelon form of $\left[\begin{array}{c|c} A & I \end{array}\right]$.

### code
set_random_seed(0)
print_ans = False
A = random_good_matrix(4,4,4)
AI = A.augment(identity_matrix(4), subdivide=True)
IB = AI.rref()
B = IB[:,4:]

print("[ A | I ] =")
show(AI)
print("[ I | B ] =")
show(IB)

Exercise 1(a)

令 $\bb_i$ 為 $B$ 的第 $i$ 個行向量。
令 $\be_i$ 為 $I$ 的第 $i$ 個行向量。
驗證是否 $A\bb_i = \be_i$。
說明為什麼。

Let $\bb_i$ be the $i$-th column of $B$ and $\be_i$ the $i$-th column of $I$. Check if $A\bb_i = \be_i$. Can you explain it?

Exercise 1(b)

驗證是否 $AB = I$。

Verify that $AB = I$.

Exercises

Exercise 2

令 $A$ 和 $B$ 為 $n\times n$ 矩陣。
驗證以下反矩陣的性質。

Let $A$ and $B$ be $n\times n$ matrices. Verify the following properties of an inverse matrix.

Exercise 2(a)

若 $A$ 可逆。
如果 $X$ 和 $B$ 是 $n\times m$ 矩陣且 $AX = B$
則 $X = A^{-1}B$。
如果 $X$ 和 $B$ 是 $m\times n$ 矩陣且 $XA = B$
則 $X = BA^{-1}$。

Suppose $A$ is invertible. If $X$ and $B$ are $n\times m$ matrices and $AX = B$, then $X = A^{-1}B$. If $X$ and $B$ are $m\times n$ matrices and $XA = B$, then $X = BA^{-1}$.

Exercise 2(b)

若 $A$ 和 $B$ 都可逆﹐
則 $AB$ 也可逆且 $(AB)^{-1} = B^{-1}A^{-1}$。

If $A$ and $B$ are invertible, then $AB$ is also invertible with $(AB)^{-1} = B^{-1}A^{-1}$.

Exercise 2(c)

若 $AB$ 可逆﹐
則 $A$ 和 $B$ 都可逆。

If $AB$ are invertible, then both $A$ and $B$ are invertible.

Exercise 2(d)

若 $A$ 可逆且 $B = A^{-1}$﹐
則 $B$ 也可逆且 $A = B^{-1}$。

If $A$ is invertible with $B = A^{-1}$, then $B$ is invertible with $A = B^{-1}$.

Exercise 2(e)

若 $A$ 可逆﹐
則 $A\trans$ 也可逆且 $(A\trans)^{-1} = (A^{-1})\trans$。

If $A$ is invertible, then $A\trans$ is invertible with $(A\trans)^{-1} = (A^{-1})\trans$.

Exercise 3

令 $A$ 是一個 $m\times n$ 矩陣。
以下討論 $A\trans A$ 是否可逆。

Let $A$ be an $m\times n$. The following problems are about the invertibility of $A\trans A$.

Exercise 3(a)

證明以下敘述等價：

1. $\ker(A) = \{\bzero\}$（因此必有 $m\geq n$）。
2. $A\trans A$ 可逆。

Show that the following are equivalent:

1. $\ker(A) = \{\bzero\}$ (so necessarily $m\geq n$).
2. $A\trans A$ is invertible.

Exercise 3(b)

證明若 $m < n$ 則 $A\trans A$ 不可逆。

Show that if $m < n$ then $A\trans A$ is not invertible.

Exercise 4

令 $A$ 和 $B$ 為 $n\times n$ 矩陣。
依照以下步驟證明
若 $AB = I_n$ 則 $BA = I_n$。

Let $A$ and $B$ be $n\times n$ matrices. Use the given instructions to prove that $AB = I_n$ if and only if $BA = I_n$.

Exercise 4(a)

若 $AB = I_n$。
證明增廣矩陣 $\left[\begin{array}{c|c} A & I_n \end{array}\right]$ 的最簡階梯形式矩陣
一定是 $\left[\begin{array}{c|c} I_n & B \end{array}\right]$。

Suppose $AB = I_n$. Show that the reduced echelon form of $\left[\begin{array}{c|c} A & I_n \end{array}\right]$ is $\left[\begin{array}{c|c} I_n & B \end{array}\right]$.

Exercise 4(b)

令 $\be_i$ 為 $I_n$ 的第 $i$ 個行向量。
令 $\ba_i$ 為 $A$ 的第 $i$ 個行向量。

說明對每個 $i = 1,\ldots, n$ 都有

$$ \left[\begin{array}{c|c} A & I_n \end{array}\right] \begin{bmatrix} \be_i \\ -\ba_i \end{bmatrix} = \bzero, $$

因此也有

$$ \left[\begin{array}{c|c} I_n & B \end{array}\right] \begin{bmatrix} \be_i \\ -\ba_i \end{bmatrix} = \bzero. $$

Let $\be_i$ be the $i$-th column of $I_n$ and $\ba_i$ the $i$-th column of $A$.

Show that for each $i = 1,\ldots, n$, we have

$$ \left[\begin{array}{c|c} A & I_n \end{array}\right] \begin{bmatrix} \be_i \\ -\ba_i \end{bmatrix} = \bzero, $$

and

$$ \left[\begin{array}{c|c} I_n & B \end{array}\right] \begin{bmatrix} \be_i \\ -\ba_i \end{bmatrix} = \bzero. $$

Exercise 4(c)

因為對每個 $i = 1,\ldots, n$ 都有

$$ \left[\begin{array}{c|c} I_n & B \end{array}\right] \begin{bmatrix} \be_i \\ -\ba_i \end{bmatrix} = \bzero. $$

說明對每個 $i = 1,\ldots, n$ 都有 $B\ba_i = \be_i$、
因此 $BA = I_n$。

For each $i = 1,\ldots, n$, we know

$$ \left[\begin{array}{c|c} I_n & B \end{array}\right] \begin{bmatrix} \be_i \\ -\ba_i \end{bmatrix} = \bzero. $$

Use this fact to show that $B\ba_i = \be_i$ for each $i = 1,\ldots, n$. Therefore, $BA = I_n$.

Exercise 4(d)

找一組例子使得
$A$ 是 $n\times m$ 矩陣、
$B$ 是 $m\times n$ 矩陣、
$AB = I_n$、
但是 $BA \neq I_m$。

Find an $n\times m$ matrix $A$ and an $m\times n$ matrix $B$ such that $AB = I_n$ but $BA \neq I_m$.

Exercise 5

令 $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ 且 $\det(A) \neq 0$。
證明 $A^{-1} = \frac{1}{\det(A)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$。

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ with $\det(A) \neq 0$. Show that $A^{-1} = \frac{1}{\det(A)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.