第 05 章

转置、排列与向量空间 R^n

转置翻转行列，排列矩阵重排行序，二者都满足「反序」规则；向量空间则要求对加法与数乘封闭，这个要求直接推出「必须包含零向量」。

Transposes, Permutations, and the Vector Space R^n

Transposes flip rows and columns, permutation matrices reorder rows, and both obey a reversal rule. A vector space is closed under addition and scalar multiplication — which forces it to contain the zero vector.

首发于 2009.05.07中 / EN

系统课程linear-algebratransposepermutationvector-spacesubspace

篇内目录

本讲的三件事

排列矩阵：有行交换时的消元怎么写。
转置与对称矩阵：一个几乎免费就能造出对称矩阵的方法。
向量空间与子空间：从「一堆向量」升级到「一个空间」。

前两件是上一讲的收尾，第三件是整门课真正开始的地方。

时间轴

时间	内容
0:21	排列矩阵与带行交换的消元
2:06	交换行得到主元
3:44	MATLAB 如何记录行交换
9:13	排列矩阵的逆
10:33	转置的例子
12:52	对称矩阵
15:44	$R^TR$ 总是对称
18:27	转置规则：顺序反转
21:17	向量加法与线性组合
26:54	$R^3$ 中的向量
32:32	子空间的定义
36:03	$R^2$ 的所有子空间
42:46	从矩阵造出列空间
44:58	列空间是平面
47:22	其他子空间

排列矩阵与带行交换的消元

消元时如果主元位置出现 $0$ ，就要交换行。排列矩阵 $P$ 就是把单位矩阵的行重新排列得到的矩阵；单位矩阵本身也算一个排列（什么都不做）。

$n\times n$ 的排列矩阵一共有 $n!$ 个： $3\times3$ 有 $6$ 个， $4\times4$ 有 $24$ 个， $5\times5$ 有 $120$ 个。

它们构成一个群：

两个排列相乘，仍是排列；
每个排列都可逆；
逆就是转置：

P^T P=I \;\Longrightarrow\; P^{-1}=P^T.

验证的思路很直接：交换两行再做一次就还原，所以很多 $P$ 是自己的逆；对更长的循环，转置恰好把「谁移到哪里」倒过来。

有行交换时，消元写作

PA=LU.

$P$ 先记录行的正确顺序， $L$ 与 $U$ 的结构不变。

补充： 实际数值计算中，MATLAB 一类软件不只检查主元是否为零，还会检查它是否足够大。主元非常接近零会放大舍入误差，所以软件会做「部分主元法」：在每一列挑选绝对值最大的元素换到主元位置。代数上不必要，数值上必要。

转置

定义与例子

转置把行和列互换：

(A^T)_{ij}=A_{ji}.

例如

R=\begin{bmatrix}1&2&4\\3&3&1\end{bmatrix}, \qquad R^T=\begin{bmatrix}1&3\\2&3\\4&1\end{bmatrix}.

$R$ 是 $2\times3$ ， $R^T$ 是 $3\times2$ 。

乘积的转置也反序

(AB)^T=B^T A^T.

和逆的规则一样，顺序要反过来。

对称矩阵

如果 $A^T=A$ ，就称 $A$ 对称。对称矩阵关于主对角线镜像相等。

一个几乎免费得到对称矩阵的办法：对任意矩阵 $R$ ，乘积 $R^T R$ 一定对称。证明只有两行：

(R^T R)^T=R^T (R^T)^T=R^T R.

用上面的 $R$ 计算：

R^T R= \begin{bmatrix} 1&3\\2&3\\4&1 \end{bmatrix} \begin{bmatrix}1&2&4\\3&3&1\end{bmatrix} = \begin{bmatrix} 10&11&7\\ 11&13&11\\ 7&11&17 \end{bmatrix}.

第 $(1,2)$ 项是 $1\cdot2+3\cdot3=11$ ，第 $(2,1)$ 项是 $2\cdot1+3\cdot3=11$ ，相等不是巧合。

一句话总结： 对称性不是碰巧，而是「转置两次回到原矩阵」这条规则的必然结果。

向量空间：先看运算，再看集合

两种基本运算

线性代数只做两件事：

向量相加；
向量乘标量。

如果一组向量对这两种运算都封闭——加完、乘完仍然在这组里——并且满足通常的运算律，它就是一个向量空间。

$R^2$ 与 $R^n$

$R^2$ 是所有二元实向量组成的空间，几何上就是整个平面。 $R^3$ 是三维空间。 $R^n$ 是所有 $n$ 个实分量向量组成的空间。

约定：本课程使用列向量。

为什么零向量必须在

考虑 $R^2$ 中「去掉原点」的平面。它不是向量空间，因为：

取向量 $[3,2]^T$ ，乘以 $0$ ，结果 $[0,0]^T$ 不在集合里；
取 $[3,2]^T$ 与 $[-3,-2]^T$ ，相加也回到原点，同样不在集合里。

所以每个向量空间都必须包含零向量。这不是额外规定，而是「对数乘封闭」的直接推论。

反例：第一象限

只取坐标非负的向量，加法没问题： $[3,2]+[5,6]=[8,8]$ 仍在第一象限。但乘以 $-5$ 就跑到第三象限，封闭性被破坏。因此它不是向量空间。

子空间

子空间是包含在更大空间里、自身也满足向量空间规则的集合。

$R^2$ 的全部子空间

只有三类：

整个 $R^2$ ；
过原点的直线；
只含零向量 $\{\mathbf{0}\}$ 。

不过原点的直线不是子空间：把线上一个向量乘以 $0$ ，就离开了这条线。

注意「 $R^2$ 中过原点的直线」与 $R^1$ 很像，但不相同：前者的向量有两个分量，后者只有一个。

$R^3$ 的全部子空间

四类：

整个 $R^3$ ；
过原点的平面；
过原点的直线；
$\{\mathbf{0}\}$ 。

维度从 3 到 0，逐级下降。

从矩阵造出子空间：列空间

取矩阵

A=\begin{bmatrix} 1&3\\ 2&3\\ 4&1 \end{bmatrix}.

它的两列 $\mathbf{c}_1=[1,2,4]^T$ 、 $\mathbf{c}_2=[3,3,1]^T$ 都是 $R^3$ 中的向量。

只把这两列放在一起，还不构成子空间——必须把它们的所有线性组合都加进来：

\{x_1\mathbf{c}_1+x_2\mathbf{c}_2\mid x_1,x_2\in R\}.

这个集合叫 $A$ 的列空间，记作 $C(A)$ 。

几何上，两个不共线的向量张成一个过原点的平面。所以这个例子的列空间是 $R^3$ 中的一个平面。

如果两列恰好共线，列空间就退化成一条直线。子空间的大小取决于列向量之间的独立性——这正是下一讲要量化的东西。

推广到高维

$R^{10}$ 中取五个向量，它们的组合形成某个子空间。它不可能是 $R^5$ ——向量有十个分量——而是一个过原点的「五维平坦对象」（如果五个向量线性无关的话）；如果它们共线，就只剩一条线。

课程在这里的提醒：你必须在 $R^3$ 里真正「看见」两个向量张成一个平面，之后才能在 $R^{10}$ 里相信同样的结论。

本讲结论

排列矩阵是重排了行的单位矩阵，共 $n!$ 个，满足 $P^{-1}=P^T$ 。
有行交换时消元写作 $PA=LU$ 。
$(AB)^T=B^T A^T$ ，顺序反过来。
$R^T R$ 永远对称；对称矩阵满足 $A^T=A$ 。
向量空间对加法与数乘封闭，因此必含零向量。
子空间是更大空间内部、自身也是向量空间的集合。
$R^2$ 的子空间只有：全空间、过原点直线、零空间。
矩阵的列空间由列的所有线性组合构成；它一定是子空间。

理解检验

1. 为什么排列矩阵的逆等于它的转置？用「交换两次还原」解释。

参考答案：排列矩阵做的动作是「把行按某个置换搬动一次」。逆矩阵要做的就是把这次搬动撤销，也就是按相反的置换搬回去。交换两行是最基本的情形：换一次再换一次就回到原样，所以单个交换矩阵满足 $P^2=I$ ，而它恰好对称（ $P^T=P$ ），于是 $P^{-1}=P=P^T$ 。一般的排列可以看成若干次交换的复合，转置把复合的顺序倒过来，正好完成撤销。形式化的验证是 $P^TP=I$ ： $P$ 的列是互相垂直的标准基向量， $P^T$ 的第 $i$ 行与 $P$ 的第 $j$ 列点积，等于 $1$ 当且仅当 $i=j$ 。

2. 写出 $R^T R$ 对称的两行证明，并指出哪一步用了 $(AB)^T=B^TA^T$ 。

参考答案：

(R^T R)^T=R^T(R^T)^T=R^T R .

第一步用了乘积转置的反序规则 $(AB)^T=B^TA^T$ ，其中 $A=R^T$ 、 $B=R$ ；第二步用了「转置两次回到原矩阵」 $(R^T)^T=R$ 。于是 $(R^TR)^T=R^TR$ ，按定义这就是对称。以文中的 $R$ 为例， $R^TR=\begin{bmatrix}10&11&7\\11&13&11\\7&11&17\end{bmatrix}$ ， $(1,2)$ 与 $(2,1)$ 都是 $1\cdot2+3\cdot3=11$ ，相等不是巧合。

3. 第一象限为什么不是 $R^2$ 的子空间？给出一个具体的反例。

参考答案：子空间要对数乘封闭，第一象限做不到。取 $\mathbf{v}=[3,2]^T$ ，它坐标非负，在第一象限；取标量 $c=-5$ ，则

c\mathbf{v}=\begin{bmatrix}-15\\-10\end{bmatrix},

两个分量都变成负数，落在第三象限，已经离开了第一象限。所以第一象限只对加法（和非负数乘）封闭，对一般的数乘不封闭，不是子空间。顺带一提，它还不含零向量之外的很多「反向」向量，这也说明它不是向量空间。

4. 不过原点的直线为什么不是子空间？

参考答案：子空间必须包含零向量，而且要对数乘封闭。设直线 $L$ 不过原点，取线上任一向量 $\mathbf{v}\ne\mathbf{0}$ ，则 $0\cdot\mathbf{v}=\mathbf{0}$ 。如果 $L$ 是子空间，零向量就必须在 $L$ 上，但 $L$ 不过原点，矛盾。几何上，把线上一点沿该方向缩到 $0$ 倍，就跳出了这条线。同理，两条平行但不过原点的直线相加也可能跑出集合外。

5. 对 $A=\begin{bmatrix}1&3\\2&3\\4&1\end{bmatrix}$ ，解释为什么它的列空间是一个平面而不是整条 $R^3$ 。

参考答案： $A$ 只有两列 $\mathbf{c}_1=[1,2,4]^T$ 与 $\mathbf{c}_2=[3,3,1]^T$ ，它们不是彼此的倍数（对应分量之比 $3,\ 3/2,\ 1/4$ 不相同），所以线性无关。两个无关向量在 $R^3$ 中张成一个二维子空间，即过原点的平面。它不可能是整个 $R^3$ ： $R^3$ 需要三个方向，而两列的所有线性组合 $x_1\mathbf{c}_1+x_2\mathbf{c}_2$ 只有两个自由度，平面外的向量无论怎么组合都到不了。列空间的维数等于独立列的个数，这里是 2。

与前后课的关系

这一讲建立了「空间」的语言。下一讲用它重写 $A\mathbf{x}=\mathbf{b}$ ：有解当且仅当 $\mathbf{b}$ 落在列空间里；同时引入零空间，作为另一个由矩阵自然产生的子空间。

Three topics

Permutation matrices: how to write elimination when rows must be exchanged.
Transposes and symmetric matrices: an almost free way to build a symmetric matrix.
Vector spaces and subspaces: from "a bunch of vectors" to "a space".

The first two finish the previous lecture. The third is where the course properly begins.

Timeline

Time	Topic
0:21	Permutations and elimination with row exchanges
2:06	Exchanging rows to get a pivot
3:44	How MATLAB records row exchanges
9:13	The inverse of a permutation matrix
10:33	An example of a transpose
12:52	Symmetric matrices
15:44	$R^TR$ is always symmetric
18:27	The transpose rule: reverse the order
21:17	Vectors, addition and linear combinations
26:54	Vectors in $R^3$
32:32	The definition of a subspace
36:03	All subspaces of $R^2$
42:46	The column space of a matrix
44:58	The column space as a plane
47:22	Other subspaces

Permutations and elimination with row exchanges

If a zero appears in a pivot position, rows must be swapped. A permutation matrix $P$ is the identity with its rows rearranged; the identity itself counts as a permutation (the one that does nothing).

There are $n!$ permutation matrices of size $n$ : six of size $3$ , twenty-four of size $4$ , one hundred twenty of size $5$ .

They form a group:

the product of two permutations is a permutation;
every permutation is invertible;
the inverse is the transpose:

P^T P=I \;\Longrightarrow\; P^{-1}=P^T.

The intuition: swapping two rows twice restores the original, so many permutations are their own inverse; for longer cycles, transposing reverses "who moves where".

With row exchanges, elimination is written

PA=LU.

$P$ records the correct row order, and $L$ and $U$ keep their structure.

Background: In actual numerical software, MATLAB does not merely test whether a pivot is zero; it tests whether it is large enough. A pivot very close to zero amplifies rounding error, so the software uses partial pivoting: it moves the largest-magnitude entry in the column into the pivot position. Algebra does not require this; numerical accuracy does.

Transposes

Definition and example

The transpose swaps rows and columns:

(A^T)_{ij}=A_{ji}.

For example,

R=\begin{bmatrix}1&2&4\\3&3&1\end{bmatrix}, \qquad R^T=\begin{bmatrix}1&3\\2&3\\4&1\end{bmatrix}.

$R$ is $2\times3$ ; $R^T$ is $3\times2$ .

The transpose of a product reverses the order

(AB)^T=B^T A^T.

The same reversal rule as for inverses.

Symmetric matrices

If $A^T=A$ , the matrix is symmetric: entries mirror across the main diagonal.

There is an almost free way to produce one: for any matrix $R$ , the product $R^T R$ is always symmetric. The proof takes two lines:

(R^T R)^T=R^T (R^T)^T=R^T R.

With the $R$ above:

R^T R= \begin{bmatrix} 1&3\\2&3\\4&1 \end{bmatrix} \begin{bmatrix}1&2&4\\3&3&1\end{bmatrix} = \begin{bmatrix} 10&11&7\\ 11&13&11\\ 7&11&17 \end{bmatrix}.

Entry $(1,2)$ is $1\cdot2+3\cdot3=11$ , and entry $(2,1)$ is $2\cdot1+3\cdot3=11$ . The equality is not a coincidence.

One-sentence summary: Symmetry is not luck; it is forced by the rule that transposing twice returns the original matrix.

Vector spaces: operations first, sets second

Two basic operations

Linear algebra does two things:

add vectors;
multiply vectors by scalars.

A collection of vectors is a vector space when it is closed under both operations — the results stay in the collection — and the usual algebraic laws hold.

$R^2$ and $R^n$

$R^2$ is all pairs of real numbers; geometrically, the whole plane. $R^3$ is three-dimensional space. $R^n$ is all vectors with $n$ real components.

Convention: the course uses column vectors.

Why zero must be included

Take the plane with the origin removed. It is not a vector space:

multiply $[3,2]^T$ by $0$ and the result $[0,0]^T$ is missing;
add $[3,2]^T$ and $[-3,-2]^T$ and you return to the origin, which is also missing.

So every vector space contains the zero vector. This is not an extra axiom; it follows from closure under scalar multiplication.

Counterexample: the first quadrant

Take only vectors with nonnegative coordinates. Addition is fine: $[3,2]+[5,6]=[8,8]$ stays in the quadrant. But multiplying by $-5$ lands in the third quadrant, so closure fails. It is not a vector space.

Subspaces

A subspace is a set inside a larger space that is itself a vector space under the same operations.

All subspaces of $R^2$

Only three:

all of $R^2$ ;
a line through the origin;
the zero vector alone, $\{\mathbf{0}\}$ .

A line not through the origin is not a subspace: multiply one of its vectors by $0$ and you leave the line.

A line through the origin in $R^2$ looks like $R^1$ , but they are not the same: vectors in the line have two components, vectors in $R^1$ have one.

All subspaces of $R^3$

Four:

all of $R^3$ ;
a plane through the origin;
a line through the origin;
$\{\mathbf{0}\}$ .

Dimensions run from 3 down to 0.

Building a subspace from a matrix: the column space

Take

A=\begin{bmatrix} 1&3\\ 2&3\\ 4&1 \end{bmatrix}.

Its columns $\mathbf{c}_1=[1,2,4]^T$ and $\mathbf{c}_2=[3,3,1]^T$ live in $R^3$ .

Two columns by themselves are not a subspace. To get one, include all linear combinations:

\{x_1\mathbf{c}_1+x_2\mathbf{c}_2\mid x_1,x_2\in R\}.

This is the column space of $A$ , written $C(A)$ .

Geometrically, two non-collinear vectors span a plane through the origin. So this column space is a plane inside $R^3$ .

If the two columns happened to be collinear, the column space would shrink to a line. How large the subspace is depends on independence among the columns — the quantity the next lecture makes precise.

Higher dimensions

Take five vectors in $R^{10}$ . Their combinations form some subspace. It cannot be $R^5$ — the vectors have ten components — but it is a five-dimensional flat object through the origin (if the five are independent); if they are all collinear, it is only a line.

The lecture's warning: you must genuinely see two vectors spanning a plane in $R^3$ before you can trust the same conclusion in $R^{10}$ .

Takeaways

A permutation matrix is the identity with reordered rows; there are $n!$ of them, and $P^{-1}=P^T$ .
With row exchanges, elimination is $PA=LU$ .
$(AB)^T=B^T A^T$ : the order reverses.
$R^T R$ is always symmetric; a symmetric matrix satisfies $A^T=A$ .
A vector space is closed under addition and scalar multiplication, hence contains zero.
A subspace is a vector space sitting inside a larger one.
The subspaces of $R^2$ are the whole space, lines through the origin, and $\{\mathbf{0}\}$ .
The column space consists of all linear combinations of the columns; it is always a subspace.

Check your understanding

1. Why is the inverse of a permutation matrix its transpose? Use "swap twice and you are back".

Reference answer: A permutation matrix rearranges rows once; its inverse must undo that rearrangement, i.e. permute back. Row swap is the basic case: swapping twice restores the original, so a single swap matrix satisfies $P^2=I$ , and it is symmetric ( $P^T=P$ ), hence $P^{-1}=P=P^T$ . A general permutation is a composite of swaps, and transposing reverses the order of the composite, which is exactly what undoing requires. Formally $P^TP=I$ : the columns of $P$ are mutually orthogonal standard basis vectors, so the dot product of row $i$ of $P^T$ with column $j$ of $P$ is $1$ only when $i=j$ .

2. Write the two-line proof that $R^T R$ is symmetric and identify where $(AB)^T=B^T A^T$ is used.

Reference answer:

(R^T R)^T=R^T(R^T)^T=R^T R .

The first step is the reversed-order rule $(AB)^T=B^TA^T$ with $A=R^T$ and $B=R$ ; the second uses $(R^T)^T=R$ . Hence $(R^TR)^T=R^TR$ , which is symmetry by definition. For the $R$ in the lecture, $R^TR=\begin{bmatrix}10&11&7\\11&13&11\\7&11&17\end{bmatrix}$ , and entries $(1,2)$ and $(2,1)$ are both $1\cdot2+3\cdot3=11$ — not a coincidence.

3. Why is the first quadrant not a subspace of $R^2$ ? Give a concrete counterexample.

Reference answer: A subspace must be closed under scalar multiplication, and the first quadrant is not. Take $\mathbf{v}=[3,2]^T$ , which lies in the quadrant, and $c=-5$ :

c\mathbf{v}=\begin{bmatrix}-15\\-10\end{bmatrix},

both coordinates negative, in the third quadrant. The quadrant is closed under addition and nonnegative scaling only; a general scalar breaks closure, so it is not a subspace.

4. Why is a line not through the origin not a subspace?

Reference answer: A subspace must contain the zero vector. Take any $\mathbf{v}\ne\mathbf{0}$ on a line $L$ that misses the origin; then $0\cdot\mathbf{v}=\mathbf{0}$ , so if $L$ were a subspace the origin would have to lie on it — contradiction. Geometrically, scaling a point on the line down to zero jumps off the line.

5. For $A=\begin{bmatrix}1&3\\2&3\\4&1\end{bmatrix}$ , explain why its column space is a plane rather than all of $R^3$ .

Reference answer: $A$ has only two columns, $\mathbf{c}_1=[1,2,4]^T$ and $\mathbf{c}_2=[3,3,1]^T$ , and they are not multiples of each other (the ratios $3,\ 3/2,\ 1/4$ differ), so they are independent. Two independent vectors in $R^3$ span a two-dimensional subspace, a plane through the origin. It cannot be all of $R^3$ : reaching every point needs three directions, while $x_1\mathbf{c}_1+x_2\mathbf{c}_2$ has only two degrees of freedom, so vectors off the plane are unreachable.

How it connects

This lecture builds the language of spaces. The next one uses it to rewrite $A\mathbf{x}=\mathbf{b}$ : a solution exists exactly when $\mathbf{b}$ lies in the column space. It also introduces the nullspace, the other subspace a matrix naturally produces.

来源

视频Lecture 5: Transposes, Permutations, Spaces R^nGilbert Strang, MIT OpenCourseWare · 2009-05-07