Hefferon的线性代数教程：简明易懂的自学习资源

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线性代数是一门基础且应用广泛的数学分支，它研究向量空间、矩阵、线性变换和它们之间的关系。本教程由Jim Hefferon撰写，特别强调了英文版的学习方式，旨在使自学过程清晰易懂。以下是一些核心概念： 1. **向量空间** (Vector Spaces)：在数学中，向量空间是满足特定性质的一组对象集合，通常包括实数（R）、自然数（N）、复数（C）等。向量空间V中的元素（如向量⃗ v 和⃗ w）具有加法和标量乘法运算，以及零向量（⃗ 0V）的概念。 2. **基与标准基** (Bases and Standard Basis)：基（B）是一组线性无关且生成整个向量空间的向量集合。例如，在实数向量空间Rn中，标准基En由n个单位向量组成（⃗ e1, ⃗ e2, ..., ⃗ en）。 3. **向量表示** (Vector Representation)：任何向量⃗ v 可以通过其在给定基B下的系数矩阵（RepB(⃗ v)）来表示，这体现了向量在不同基之间的转换。 4. **矩阵** (Matrices)：矩阵用于表示线性变换，如矩阵H和G代表线性映射。矩阵的行和列对应于向量空间的维数，矩阵元素hi,j代表从第i行到第j列的映射。 5. **线性变换** (Linear Transformations)：t和s是线性变换或映射，它们将一个向量空间映射到另一个向量空间。例如，T和S是方阵，它们的行列式（|T|）是衡量其秩的重要指标。 6. **子空间与直和** (Subspaces and Direct Sum)：Mn×m表示n×m的矩阵集合，子空间是向量空间内的特定集合。M⊕N表示两个子空间的直接和，即这两个子空间中的向量组合在一起形成一个新的子空间。 7. **同构空间** (Isomorphic Spaces)：当两个向量空间在某种意义上可以相互替代，即存在一个一对一的映射使得它们的结构保持不变时，称它们是同构的。 8. **核与像** (Range and Null Space)：R(h)和N(h)分别指线性映射h的像和核，前者是映射作用后的所有可能结果集合，后者是不被映射到任何非零元素的向量集合。 9. **泛函分析概念** (Generalized Concepts)：R∞(h)和N∞(h)分别指代广义的像和核，用于处理无限维度的向量空间和映射。 10. **希腊字母符号** (Greek Alphabet)：文中使用的希腊字母，如α、β、γ等，通常用于表示变量、指数、元素或特定的数学概念，比如向量的分量或矩阵的行列式。掌握这些概念有助于理解线性代数的基本理论和应用，无论是解决线性方程组、特征值问题，还是在数据科学、机器学习等领域，线性代数都是不可或缺的基础工具。

6 Chapter 1. Linear Systems

the second equation has no leading y. To get one, we look lower down in the

system for a row that has a leading y and swap it in.

↔ρ

−→

x − y =0

y + w =0

z +2w =4

2z + w =5

(Had there been more than one row below the second with a leading y then we

could have swapped in any one.) The rest of Gauss’ method goes as before.

−2ρ

+ρ

−→

x − y =0

y + w =0

z +2w =4

−3w = −3

Back-substitution gives w =1,z =2,y = −1, and x = −1.

Strictly speaking, the operation of rescaling rows is not needed to solve linear

systems. We have included it because we will use it later in this chapter as part

of a variation on Gauss’ method, the Gauss-Jordan method.

All of the systems seen so far have the same number of equations as un-

knowns. All of them have a solution, and for all of them there is only one

solution. We ﬁnish this subsection by seeing for contrast some other things that

can happen.

1.11 Example Linear systems need not have the same number of equations

as unknowns. This system

x +3y =1

2x + y = −3

2x +2y = −2

has more equations than variables. Gauss’ method helps us understand this

system also, since this

−2ρ

+ρ

−→

−2ρ

+ρ

x +3y =1

−5y = −5

−4y = −4

shows that one of the equations is redundant. Echelon form

−(4/5)ρ

+ρ

−→

x +3y =1

−5y = −5

0= 0

gives y =1andx = −2. The ‘0 = 0’ is derived from the redundancy.

8 Chapter 1. Linear Systems

Don’t be fooled by the ‘0 = 0’ equation in that example. It is not the signal

that a system has many solutions.

1.15 Example The absence of a ‘0 = 0’ does not keep a system from having

many diﬀerent solutions. This system is in echelon form

x + y + z =0

y + z =0

has no ‘0 = 0’, and yet has inﬁnitely many solutions. (For instance, each of

these is a solution: (0, 1, −1), (0, 1/2, −1/2), (0, 0, 0), and (0, −π, π). There are

inﬁnitely many solutions because any triple whose ﬁrst component is 0 and

whose second component is the negative of the third is a solution.)

Nor does the presence of a ‘0 = 0’ mean that the system must have many

solutions. Example 1.11 shows that. So does this system, which does not have

many solutions — in fact it has none — despite that when it is brought to

echelon form it has a ‘0 = 0’ row.

2x − 2z =6

y + z =1

2x + y − z =7

3y +3z =0

−ρ

+ρ

−→

2x −2z =6

y + z =1

3y +3z =0

−ρ

+ρ

−→

−3ρ

+ρ

2x −2z =6

y + z =1

0= 0

0=−3

We will ﬁnish this subsection with a summary of what we’ve seen so far

about Gauss’ method.

Gauss’ method uses the three row operations to set a system up for back

substitution. If any step shows a contradictory equation then we can stop

with the conclusion that the system has no solutions. If we reach echelon form

without a contradictory equation, and each variable is a leading variable in its

row, then the system has a unique solution and we ﬁnd it by back substitution.

Finally, if we reach echelon form without a contradictory equation, and there is

not a unique solution (at least one variable is not a leading variable) then the

system has many solutions.

The next subsection deals with the third case — we will see how to describe

the solution set of a system with many solutions.

Exercises

X 1.16 Use Gauss’ method to ﬁnd the unique solution for each system.

(a)

2x +3y =13

x − y = −1

(b)

x − z =0

3x + y =1

−x + y + z =4

X 1.17 Use Gauss’ method to solve each system or conclude ‘many solutions’ or ‘no

solutions’.

Section I. Solving Linear Systems 9

(a) 2x +2y =5

x − 4y =0

(b) −x + y =1

x + y =2

x + y +2z =14

(d) −x − y =1

−3x − 3y =2

(e) 4y + z =20

2x − 2y + z =0

x + z =5

x + y − z =10

(f) 2x + z + w =5

y − w = −1

3x − z − w =0

4x + y +2z + w =9

X 1.18 There are methods for solving linear systems other than Gauss’ method. One

often taught in high school is to solve one of the equations for a variable, then

substitute the resulting expression into other equations. That step is repeated

until there is an equation with only one variable. From that, the ﬁrst number in

the solution is derived, and then back-substitution can be done. This method both

takes longer than Gauss’ method, since it involves more arithmetic operations and

is more likely to lead to errors. To illustrate how it can lead to wrong conclusions,

we will use the system

x +3y =1

2x + y = −3

2x +2y =0

from Example 1.12.

(a) Solve the ﬁrst equation for x and substitute that expression into the second

equation. Find the resulting y.

(b) Again solve the ﬁrst equation for x, but this time substitute that expression

into the third equation. Find this y.

What extra step must a user of this method take to avoid erroneously concluding

a system has a solution?

X 1.19 For which values of k are there no solutions, many solutions, or a unique

solution to this system?

x − y =1

3x − 3y = k

X 1.20 This system is not linear:

2sinα − cos β +3tanγ =3

4sinα +2cosβ − 2tanγ =10

6sinα − 3cosβ +tanγ =9

but we can nonetheless apply Gauss’ method. Do so. Does the system have a

solution?

X 1.21 What conditions must the constants, the b’s, satisfy so that each of these

systems has a solution? Hint. Apply Gauss’ method and see what happens to the

right side.

(a) x − 3y = b

3x + y = b

x +7y = b

2x +4y = b

(b) x

+2x

+3x

= b

+5x

+3x

= b

+8x

= b

1.22 True or false: a system with more unknowns than equations has at least one

solution. (As always, to say ‘true’ you must prove it, while to say ‘false’ you must

produce a counterexample.)

1.23 Must any Chemistry problem like the one that starts this subsection — a

balance the reaction problem — have inﬁnitely many solutions?

X 1.24 Find the coeﬃcients a, b,andc so that the graph of f(x)=ax

+bx+c passes

through the points (1, 2), (−1, 6), and (2, 3).

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