本文为《Linear algebra and its applications》的读书笔记
目录
Vectors in R 2 \R^2 R2
- A matrix with only one column is called a column vector (列向量), or simply a vector (向量).
w = [ w 1 w 2 ] \boldsymbol w=\begin{bmatrix}w_1\\w_2\end{bmatrix} w=[w1w2]where w 1 w_1 w1 and w 2 w_2 w2 are any real numbers. - The set of all vectors with two entries is denoted by R 2 \mathbb{R}^2 R2(read “r-two”). The R \mathbb{R} R stands for the real numbers that appear as entries in the vectors, and the exponent 2 indicates that each vector contains two entries. Vectors in R 2 \mathbb{R}^2 R2 are ordered pairs of real numbers. (实数的有序对)
Sometimes, for convenience (and also to save space), this text may write a column vector in the form ( w 1 , w 2 ) (w_1, w_2) (w1,w2). In this case, the parentheses and the comma distinguish the vector ( w 1 , w 2 ) (w_1, w_2) (w1,w2) from the 1 × 2 1\times 2 1×2 row matrix [ w 1 w 2 ] [w_1\ w_2] [w1 w2], written with brackets and no comma.
scalar multiple (标量乘法 / 数乘):
Geometric Descriptions of R 2 \R^2 R2
- Consider a rectangular coordinate system (直角坐标系) in the plane. We can identify a geometric point
(
a
,
b
)
(a, b)
(a,b) with the column vector
[
a
b
]
\begin{bmatrix}a\\b\end{bmatrix}
[ab].
- The sum of two vectors has a useful geometric representation. The following rule can be verified by analytic geometry. (解析几何)
For simplicity of notation, a vector such as u + ( − 1 ) v \boldsymbol u +(-1)\boldsymbol v u+(−1)v is often written as u − v \boldsymbol u -\boldsymbol v u−v.
Vectors in R n \mathbb{R}^n Rn
u = [ u 1 u 2 . . . u n ] \boldsymbol u=\begin{bmatrix}u_1\\u_2\\...\\u_n\end{bmatrix} u=⎣⎢⎢⎡u1u2...un⎦⎥⎥⎤
- The vector whose entries are all zero is called the zero vector and is denoted by 0. (The number of entries in 0 will be clear from the context.)
Linear Combinations
线性组合
Linear Combinations
y = c 1 v 1 + . . . + c p v p y=c_1\boldsymbol v_1+...+c_p\boldsymbol v_p y=c1v1+...+cpvpis called a linear combination of v 1 , . . . , v p \boldsymbol v_1,..., \boldsymbol v_p v1,...,vp with weights c 1 , . . . , c p c_1,...,c_p c1,...,cp.
The weights in a linear combination can be any real numbers
线性组合与“存在”问题的联系
- The next example connects a problem about linear combinations to the fundamental existence question studied in Sections 1.1 and 1.2.
EXAMPLE 4
Let a 1 = [ 1 − 2 − 5 ] , a 2 = [ 2 5 6 ] \boldsymbol a_1=\begin{bmatrix}1\\-2\\-5\end{bmatrix},\ \boldsymbol a_2=\begin{bmatrix}2\\5\\6\end{bmatrix} a1=⎣⎡1−2−5⎦⎤, a2=⎣⎡256⎦⎤, and b = [ 7 4 − 3 ] \boldsymbol b=\begin{bmatrix}7\\4\\-3\end{bmatrix} b=⎣⎡74−3⎦⎤. Determine whether b \boldsymbol b b can be generated (or written) as a linear combination of a 1 \boldsymbol a_1 a1 and a 2 \boldsymbol a_2 a2.
SOLUTION
- To solve this system, row reduce the augmented matrix of the system as follows:
∼ \sim ∼ 表示行等价
- Hence b \boldsymbol b b is a linear combination of a 1 \boldsymbol a_1 a1 and a 2 \boldsymbol a_2 a2, with weights x 1 = 3 x_1 = 3 x1=3 and x 2 = 2 x_2 = 2 x2=2.
- Observe in Example 5 that the original vectors
a
1
\boldsymbol a_1
a1,
a
2
\boldsymbol a_2
a2, and
b
\boldsymbol b
b are the columns of the augmented matrix that we row reduced:
For brevity, write this matrix in a way that identifies its columns—namely,
[ a 1 a 2 b ] [\ \boldsymbol a_1\ \ \ \boldsymbol a_2\ \ \ \boldsymbol b\ ] [ a1 a2 b ]
- The discussion above is easily modified to establish the following fundamental fact:
生成集 Span \text {Span} Span
- One of the key ideas in linear algebra is to study the set of all vectors that can be generated or written as a linear combination of a fixed set { v 1 , . . . , v p } \{\boldsymbol v_1,...,\boldsymbol v_p \} {v1,...,vp} of vectors.
- Asking whether a vector
b
\boldsymbol b
b is in
S
p
a
n
{
v
1
,
.
.
.
,
v
p
}
Span\{\boldsymbol v_1,...,\boldsymbol v_p \}
Span{v1,...,vp} (生成集) amounts to asking whether the vector equation
x 1 v 1 + x 2 v 2 + . . . + x p v p = b x_1\boldsymbol v_1+x_2\boldsymbol v_2+...+x_p\boldsymbol v_p=\boldsymbol b x1v1+x2v2+...+xpvp=bhas a solution, or, equivalently, asking whether the linear system with augmented matrix [ v 1 . . . v p b ] [\ \boldsymbol v_1\ \ \ ...\ \ \ \boldsymbol v_p\ \ \ \boldsymbol b\ ] [ v1 ... vp b ] has a solution.
Span { v 1 , . . . , v p } \{\boldsymbol v_1,...,\boldsymbol v_p \} {v1,...,vp} 一定包含 0 \boldsymbol 0 0
A Geometric Description of Span { v } \{\boldsymbol v\} {v} and Span { u , v } \{\boldsymbol u,\boldsymbol v \} {u,v}
- Let
v
\boldsymbol v
v and
u
\boldsymbol u
u be a nonzero vector in
R
3
\mathbb{R}^3
R3, with
v
\boldsymbol v
v not a multiple of
u
\boldsymbol u
u
- If v \boldsymbol v v is a multiple of u \boldsymbol u u, then Span { u , v } \{\boldsymbol u,\boldsymbol v \} {u,v} can be just a line through the origin. In fact, Span { u , v } \{\boldsymbol u,\boldsymbol v \} {u,v} can also be just the origin itself.
EXAMPLE 5
Let a 1 = [ 1 4 − 2 ] , a 2 = [ − 2 − 3 7 ] \boldsymbol a_1=\begin{bmatrix}1\\4\\-2\end{bmatrix},\ \boldsymbol a_2=\begin{bmatrix}-2\\-3\\7\end{bmatrix} a1=⎣⎡14−2⎦⎤, a2=⎣⎡−2−37⎦⎤, and b = [ 4 1 h ] \boldsymbol b=\begin{bmatrix}4\\1\\h\end{bmatrix} b=⎣⎡41h⎦⎤ For what value(s) of h h h is b \boldsymbol b b in the plane spanned by a 1 \boldsymbol a_1 a1 and a 2 \boldsymbol a_2 a2?
SOLUTION
- When the linear system [ a 1 a 2 b ] [\ \boldsymbol a_1\ \ \ \boldsymbol a_2\ \ \ \boldsymbol b\ ] [ a1 a2 b ] is consistent.
EXAMPLE 6
A thin triangular plate of uniform density and thickness has vertices at v 1 = ( 0 , 1 ) \boldsymbol v_1 = (0, 1) v1=(0,1), v 2 = ( 8 , 1 ) \boldsymbol v_2 = (8, 1) v2=(8,1), and v 3 = ( 2 , 4 ) \boldsymbol v_3 = (2, 4) v3=(2,4). This “balance point” of the plate coincides with the center of mass of a system consisting of three 1-gram point masses located at the vertices of the plate. Determine how to distribute a mass of 6 g at the three vertices of the plate to move the balance point of the plate to (2,2).
SOLUTION
- The condition
w
1
+
w
2
+
w
3
=
6
w_1 + w_2 + w_3 = 6
w1+w2+w3=6 and the vector equation above combine to produce a system of three equations whose augmented matrix is shown below, along with a sequence of row operations:
- Answer: Add 3.5 g at (0, 1), add .5 g at (8, 1), and add 2 g at (2, 4).
Linear Combinations in Applications
- The final example shows how scalar multiples and linear combinations can arise when a quantity (量) such as “cost” is broken down into several categories.
EXAMPLE 7
A company manufactures two products. For $1.00 worth of product B, the company spends $.45 on materials, $.25 on labor, and $.15 on overhead (管理费用). For $1.00 worth of product C, the company spends $.40 on materials, $.30 on labor, and $.15 on overhead. Let
b
=
[
.
45
.
25
.
15
]
,
c
=
[
.
40
.
30
.
15
]
\boldsymbol b=\begin{bmatrix}.45\\.25\\.15\end{bmatrix},\ \boldsymbol c=\begin{bmatrix}.40\\.30\\.15\end{bmatrix}
b=⎣⎡.45.25.15⎦⎤, c=⎣⎡.40.30.15⎦⎤Then
b
\boldsymbol b
b and
c
\boldsymbol c
c represent the “costs per dollar of income” for the two products. Suppose the company wishes to manufacture
x
1
x_1
x1 dollars worth of product B and
x
2
x_2
x2 dollars worth of product C. Give a vector that describes the various costs the company will have (for materials, labor, and overhead).
SOLUTION
x
1
b
+
x
2
c
x_1\boldsymbol b+x_2\boldsymbol c
x1b+x2c