4.2 Null spaces, column spaces, and linear transformations (零空间、列空间和线性变换)

本文为《Linear algebra and its applications》的读书笔记

In applications of linear algebra, subspaces of $\mathbb R^n$Rn usually arise in one of two ways: (1) as the set of all solutions to a system of homogeneous linear equations or (2) as the set of all linear combinations of certain specified vectors.

The Null Space of a Matrix

Consider the following system of homogeneous equations:

In matrix form, this system is written as $A\boldsymbol x = \boldsymbol 0$Ax=0, where

We call the set of $\boldsymbol x$x that satisfy $A\boldsymbol x = \boldsymbol 0$Ax=0 the null space of the matrix A.

A more dynamic description of $Nul$Nul $A$A is the set of all $\boldsymbol x$x in $\mathbb Rn$Rn that are mapped into the zero vector of $\mathbb R^m$Rm via the linear transformation $\boldsymbol x\mapsto A\boldsymbol x$x↦Ax. See Figure 1.

EXAMPLE 2
Let $H$H be the set of all vectors in $\mathbb R^4$R4 whose coordinates $a$a, $b$b, $c$c, $d$d satisfy the equations $a - 2b + 5c = d$a−2b+5c=d and $c - a = b$c−a=b. Show that $H$H is a subspace of $\mathbb R^4$R4.
SOLUTION
$H$H is the set of all solutions of the following system of homogeneous linear equations:

By Theorem 2, $H$H is a subspace of $\mathbb R^4$R4.

An Explicit Description of $Nul$Nul $A$A

There is no obvious relation between vectors in $Nul$Nul $A$A and the entries in $A$A. We say that $Nul$Nul $A$A is defined $implicitly$implicitly, because it is defined by a condition that must be checked. No explicit list or description of the elements in $Nul$Nul $A$A is given. However, $solving$solving the equation $A\boldsymbol x = \boldsymbol 0$Ax=0 amounts to producing an $explicit$explicit description of $Nul$Nul $A$A.

EXAMPLE 3
Find a spanning set for the null space of the matrix

SOLUTION
The first step is to find the general solution of $A\boldsymbol x = \boldsymbol 0$Ax=0 in terms of free variables. Row reduce the augmented matrix to reduced echelon form in order to write the basic variables in terms of the free variables:

Every linear combination of $\boldsymbol u$u, $\boldsymbol v$v, and $\boldsymbol w$w is an element of $Nul$Nul $A$A and vice versa(反之亦然). Thus $\{\boldsymbol u,\boldsymbol v,\boldsymbol w\}${u,v,w}(the explicit description) is a spanning set for $Nul\ A$Nul A.

Two points should be made about the solution of Example 3 that apply to all problems of this type where $Nul\ A$Nul A contains nonzero vectors. We will use these facts later.

1. The spanning set produced by the method in Example 3 is automatically linearly independent because the free variables are the weights on the spanning vectors. For instance, look at the 2nd, 4th, and 5th entries in the solution vector in (3) and note that $x_2\boldsymbol u + x_4\boldsymbol v + x_5\boldsymbol w$x2​u+x4​v+x5​w can be $\boldsymbol 0$0 only if the weights $x_2, x_4$x2​,x4​, and $x_5$x5​ are all zero.
2. When $Nul\ A$Nul A contains nonzero vectors, the number of vectors in the spanning set for $Nul\ A$Nul A equals the number of free variables in the equation $A\boldsymbol x = \boldsymbol 0$Ax=0.

The Column Space of a Matrix

Note that a typical vector in $Col\ A$Col A can be written as $A\boldsymbol x$Ax for some $\boldsymbol x$x because the notation $A\boldsymbol x$Ax stands for a linear combination of the columns of $A$A. That is,

The notation $A\boldsymbol x$Ax for vectors in $Col\ A$Col A also shows that $Col\ A$Col A is the range of the linear transformation $x\mapsto A\boldsymbol x$x↦Ax.

Recall that the columns of $A$A span $\mathbb R^m$Rm if and only if the equation $A\boldsymbol x = \boldsymbol b$Ax=b has a solution for each $\boldsymbol b$b. We can restate this fact as follows:

The Contrast Between $Nul\ A$Nul A and $Col\ A$Col A

EXAMPLE 5
Let

a. $Col\ A$Col A is a subspace of $\mathbb R^3$R3.
b. $Nul\ A$Nul A is a subspace of $\mathbb R^4$R4.

When a matrix is not square, as in Example 5, the vectors in $Nul\ A$Nul A and $Col\ A$Col A live in entirely different “universes.” When $A$A is square, $Nul\ A$Nul A and $Col\ A$Col A do have the zero vector in common, and in special cases it is possible that some nonzero vectors belong to both $Nul\ A$Nul A and $Col\ A$Col A.

A surprising connection between the null space and column space will emerge in Section 4.6, after more theory is available.

Kernel and Range of a Linear Transformation 线性变换的核与值域

Subspaces of vector spaces other than $\mathbb R^n$Rn are often described in terms of a linear transformation instead of a matrix. To make this precise, we generalize the definition given in Section 1.8.

The kernel (or null space) of such a $T$T is the set of all $\boldsymbol u$u in $V$V such that $T(\boldsymbol u)=\boldsymbol 0$T(u)=0 (the zero vector in $W$W).

如果把线性变换看作两个向量空间之间的同态映射，那么线性变换的核就是同态映射的同态核

The range of $T$T is the set of all vectors in $W$W of the form $T(\boldsymbol x)$T(x) for some $\boldsymbol x$x in $V$V .

If $T$T happens to arise as a matrix transformation—say, $T(\boldsymbol x)=A\boldsymbol x$T(x)=Ax for some matrix $A$A—then the kernel and the range of $T$T are just the null space and the column space of $A$A.

It is not difficult to show that the kernel of $T$T is a subspace of $V$V. Also, the range of $T$T is a subspace of $W$W.

In applications, a subspace usually arises as either the kernel or the range of an appropriate linear transformation.

For instance, the set of all solutions of a homogeneous linear differential equation(微分方程) turns out to be the kernel of a linear transformation. Typically, such a linear transformation is described in terms of one or more derivatives of a function. To explain this in any detail would take us too far afield at this point. So we consider only two examples. The first explains why the operation of differentiation is a linear transformation.

EXAMPLE 8
Let $V$V be the vector space of all real-valued functions $f$f defined on an interval $[a, b]$[a,b] with the property that they are differentiable(可导) and their derivatives are continuous functions on $[a, b]$[a,b]. Let $W$W be the vector space $C[a, b]$C[a,b] of all continuous functions on $[a, b]$[a,b], and let $D: V \rightarrow W$D:V→W be the transformation that changes $f$f in $V$V into its derivative $f'$f′. In calculus, two simple differentiation rules(微分法则) are

That is, $D$D is a linear transformation. It can be shown that the kernel of $D$D is the set of constant functions on $[a, b]$[a,b] and the range of $D$D is the set $W$W of all continuous functions on $[a, b]$[a,b].

EXAMPLE 9
Define $T : \mathbb P^2 \rightarrow R^2$T:P2→R2 by $T (\boldsymbol p)=\begin{bmatrix} \boldsymbol p_0 \\ \boldsymbol p_1 \end{bmatrix}$T(p)=[p0​p1​​]. For instance, if $\boldsymbol p(t)= 3 + 5t + 7t^2$p(t)=3+5t+7t2, then $T (\boldsymbol p)=\begin{bmatrix} 3 \\ 15 \end{bmatrix}$T(p)=[315​]. Find a polynomial $\boldsymbol p$p in $\mathbb P^2$P2 that spans the kernel of $T$T , and describe the range of $T$T .
SOLUTION
If $T(\boldsymbol p)$T(p) is the zero vector, then $\boldsymbol p(0) = 0$p(0)=0 and $\boldsymbol p(1) = 0$p(1)=0. One such polynomial is $\boldsymbol p(t) = t(t – 1)$p(t)=t(t–1). Any other quadratic polynomial that vanishes at 0 and 1 must be a multiple of $\boldsymbol p$p, so $\boldsymbol p$p spans the kernel of $T$T.
For the range of $T$T, observe that the image of the constant 1 function is $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$[11​], and the image of the polynomial $t$t is $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$[01​]. Denote these two images by $\boldsymbol u$u and $\boldsymbol v$v, respectively. Since the range
of $T$T is a subspace of $\mathbb R^2$R2 that contains $\boldsymbol u$u and $\boldsymbol v$v, the range must contain all linear combinations of $\boldsymbol u$u and $\boldsymbol v$v. By inspection, $\boldsymbol u$u and $\boldsymbol v$v are linearly independent, so they span $\mathbb R^2$R2. Thus the range of $T$T must contain all of $\mathbb R^2$R2.

EXAMPLE 10
c4s2 40.
Let $H = Span\{ \boldsymbol v_1,\boldsymbol v_2\}$H=Span{v1​,v2​} and $K = Span\{ \boldsymbol v_3,\boldsymbol v_4\}$K=Span{v3​,v4​}, where

$H$H and $K$K are planes in $\mathbb R^3$R3 through the origin, and they intersect in a line through $\boldsymbol 0$0. Find a nonzero vector $\boldsymbol w$w that generates that line.

[Hint: $\boldsymbol w$w can be written as $c_1\boldsymbol v_1 + c_2\boldsymbol v_2$c1​v1​+c2​v2​ and also as $c_3\boldsymbol v_3 + c_4\boldsymbol v_4$c3​v3​+c4​v4​. To build $\boldsymbol w$w, solve the equation $c_1\boldsymbol v_1 + c_2\boldsymbol v_2=c_3\boldsymbol v_3 + c_4\boldsymbol v_4$c1​v1​+c2​v2​=c3​v3​+c4​v4​ for the unknown $c_j ’s$cj​’s.]