吉尔伯特·斯特朗的线性代数经典教材第四版概要

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《线性代数经典教材》是吉尔伯特·斯特朗编写的第四版教材，专为国外计算机科学专业学生设计，深入浅出地讲解了线性代数的基础概念和重要理论。本书结构严谨，从矩阵与高斯消元法入手，引导读者理解线性方程组的几何表示及其求解方法。章节内容包括： 1. 矩阵与高斯消元：章节以矩阵的定义和运算为起点，介绍如何通过行变换进行高斯消元，解决线性方程组，从而理解矩阵的几何意义。 2. 向量空间与子空间：这部分深入探讨向量空间的性质，如向量线性组合、零空间和基础向量的定义，以及如何通过解线性方程来寻找特定空间的维度。 3. 线性独立与维度：学习如何判断一组向量是否线性独立，以及如何构造一个向量空间的基，这在后续理论中至关重要。 4. 四元基本子空间：进一步介绍了零空间、列空间、行空间和像空间，这些概念对于理解矩阵的作用和性质有着重要作用。 5. 图与网络：结合实际问题，通过图论中的概念解释线性代数在数据结构和网络分析中的应用。 6. 线性变换：讨论如何将线性方程的概念推广到线性映射，即矩阵在向量空间中的作用。 7. 正交性：探讨向量的正交性，包括内积、投影和最小子空间，以及如何找到正交基和快速傅里叶变换（FFT）在该领域的应用。 8. 行列式：作为矩阵的重要特征，行列式用于确定矩阵是否可逆、计算逆矩阵以及特征值等问题，本章介绍了行列式的定义、性质和相关公式。每章末尾都有详尽的复习练习，帮助读者巩固所学知识并提升解题能力。《线性代数经典教材》不仅是计算机科学专业的核心课程教材，也是工程师、科学家和数学研究者必备的参考书，它系统而全面地展示了线性代数的基本理论与实践应用。通过学习这本书，读者能够掌握线性代数的基本理论，并在解决实际问题中发挥关键作用。

1.2 The Geometry of Linear Equations 7

We use parentheses and commas when the components are listed horizontally, and

square brackets (with no commas) when a column vector is printed vertically. What

really matters is addition of vectors and multiplication by a scalar (a number). In Figure

1.4a you see a vector addition, component by component:

Vector addition



















−2

























−2







In the right-hand ﬁgure there is a multiplication by 2 (and if it had been −2 the vector

−2

b =

−2

(a) Add vectors along axes

b b

= 2

2 (column 3)

−2

−6

−2

columns 1 + 2

−1

= linear combination equals b

(b) Add columns 1 + 2 + (3 + 3)

Figure 1.4: The column picture: linear combination of columns equals b.

would have gone in the reverse direction):

Multiplication by scalars 2

























, −2



















−2

−4







Also in the right-hand ﬁgure is one of the central ideas of linear algebra. It uses both

of the basic operations; vectors are multiplied by numbers and then added. The result is

called a linear combination, and this combination solves our equation:

Linear combination 1







−2







+ 1







−6







+ 2



















−2







Equation (2) asked for multipliers u, v, w that produce the right side b. Those numbers

are u = 1, v = 1, w = 2. They give the correct combination of the columns. They also

gave the point (1,1, 2) in the row picture (where the three planes intersect).

1.2 The Geometry of Linear Equations 9

Another singular system, close to this one, has an inﬁnity of solutions. When the

6 in the last equation becomes 7, the three equations combine to give 0 = 0. Now the

third equation is the sum of the ﬁrst two. In that case the three planes have a whole line

in common (Figure 1.5c). Changing the right sides will move the planes in Figure 1.5b

parallel to themselves, and for b = (2,5, 7) the ﬁgure is suddenly different. The lowest

plane moved up to meet the others, and there is a line of solutions. Problem 1.5c is still

singular, but now it suffers from too many solutions instead of too few.

The extreme case is three parallel planes. For most right sides there is no solution

(Figure 1.5d). For special right sides (like b = (0,0,0)!) there is a whole plane of

solutions—because the three parallel planes move over to become the same.

What happens to the column picture when the system is singular? it has to go wrong;

the question is how, There are still three columns on the left side of the equations, and

we try to combine them to produce b. Stay with equation (3):

Singular case: Column picture

Three columns in the same plane

Solvable only for b in that plane













+ v













+ w













= b. (4)

For b = (2,5,7) this was possible; for b = (2,5,6) it was not. The reason is that those

three columns lie in a plane. Then every combination is also in the plane (which goes

through the origin). If the vector b is not in that plane, no solution is possible (Figure

1.6). That is by far the most likely event; a singular system generally has no solution.

But there is a chance that b does lie in the plane of the columns. In that case there are too

many solutions; the three columns can be combined in inﬁnitely many ways to produce

b. That column picture in Figure 1.6b corresponds to the row picture in Figure 1.5c.

b not in place

3 columns

in a plane

(a) no solution

b in place

3 columns

in a plane

(b) inﬁnity of solutions

Figure 1.6: Singular cases: b outside or inside the plane with all three columns.

How do we know that the three columns lie in the same plane? One answer is to ﬁnd a

combination of the columns that adds to zero. After some calculation, it is u = 3, v = 1,

w = −2. Three times column 1 equals column 2 plus twice column 3. Column 1 is in

10 Chapter 1 Matrices and Gaussian Elimination

the plane of columns 2 and 3. Only two columns are independent.

The vector b = (2,5,7) is in that plane of the columns—it is column 1 plus column

3—so (1, 0, 1) is a solution. We can add an multiple of the combination (3,−1,−2) that

gives b = 0. So there is a whole line of solutions—as we know from the row picture.

The truth is that we knew the columns would combine to give zero, because the rows

did. That is a fact of mathematics, not of computation—and it remains true in dimension

n. If the n planes have no point in common, or inﬁnitely many points, then the n

columns lie in the same plane.

If the row picture breaks down, so does the column picture. That brings out the

difference between Chapter 1 and Chapter 2. This chapter studies the most important

problem—the nonsingular case—where there is one solution and it has to be found.

Chapter 2 studies the general case, where there may be many solutions or none. In

both cases we cannot continue without a decent notation (matrix notation) and a decent

algorithm (elimination). After the exercises, we start with elimination.

Problem Set 1.2

1. For the equations x + y = 4, 2x −2y = 4, draw the row picture (two intersecting

lines) and the column picture (combination of two columns equal to the column

vector (4,4) on the right side).

2. Solve to ﬁnd a combination of the columns that equals b:

Triangular system

u − v − w = b

v + w = b

w = b

3. (Recommended) Describe the intersection of the three planes u + v + w + z = 6 and

u + w + z = 4 and u + w = 2 (all in four-dimensional space). Is it a line or a point or

an empty set? What is the intersection if the fourth plane u = −1 is included? Find

a fourth equation that leaves us with no solution.

4. Sketch these three lines and decide if the equations are solvable:

3 by 2 system

x + 2y = 2

x − y = 2

y = 1.

What happens if all right-hand sides are zero? Is there any nonzero choice of right-

hand sides that allows the three lines to intersect at the same point?

5. Find two points on the line of intersection of the three planes t = 0 and z = 0 and

x + y + z +t = 1 in four-dimensional space.

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