Chapter 1 (Linear Equations in Linear Algebra): The matrix of a linear transformation

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

目录

  • The matrix of a linear transformation
  • Geometric Linear Transformations of R 2 \R^2 R2
  • Existence and Uniqueness Questions

The matrix of a linear transformation

  • Whenever a linear transformation T T T arises geometrically or is described in words, we usually want a “formula” for T ( x ) T(\boldsymbol x) T(x). The discussion that follows shows that every linear transformation from R n \mathbb R^n Rn to R m \mathbb R^m Rm is actually a matrix transformation x ↦ A x \boldsymbol x \mapsto A\boldsymbol x xAx.
  • The key to finding A A A is to observe that T T T is completely determined by what it does to the columns of the n × n n \times n n×n identity matrix I n I_n In.

EXAMPLE 1

The columns of I 2 = [ 1 0 0 1 ] I_2=\begin{bmatrix}1&0\\0&1\end{bmatrix} I2=[1001] are e 1 = [ 1 0 ] \boldsymbol e_1=\begin{bmatrix}1\\0\end{bmatrix} e1=[10] and e 2 = [ 0 1 ] \boldsymbol e_2=\begin{bmatrix}0\\1\end{bmatrix} e2=[01]. Suppose T T T is a linear transformation from R 2 \mathbb R^2 R2 into R 3 \mathbb R^3 R3 such that T ( e 1 ) = [ 5 − 7 2 ] T(\boldsymbol e_1)=\begin{bmatrix}5\\-7\\2\end{bmatrix} T(e1)=572 and T ( e 2 ) = [ − 3 8 0 ] T(\boldsymbol e_2)=\begin{bmatrix}-3\\8\\0\end{bmatrix} T(e2)=380. With no additional information, find a formula for the image of an arbitrary x \boldsymbol x x in R 2 \mathbb R^2 R2.

SOLUTION

T ( x ) = T ( x 1 e 1 + x 2 e 2 ) = x 1 T ( e 1 ) + x 2 T ( e 2 ) = [ T ( e 1 ) T ( e 2 ) ] [ x 1 x 2 ] = A x \begin{aligned}T(\boldsymbol x)&=T(x_1\boldsymbol e_1+x_2\boldsymbol e_2)=x_1T(\boldsymbol e_1)+x_2T(\boldsymbol e_2) \\&=\begin{bmatrix}T(\boldsymbol e_1)&T(\boldsymbol e_2)\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix} \\&=A\boldsymbol x\end{aligned} T(x)=T(x1e1+x2e2)=x1T(e1)+x2T(e2)=[T(e1)T(e2)][x1x2]=Ax


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  • The matrix A A A in (3) is called the standard matrix for the linear transformation T T T.

Geometric Linear Transformations of R 2 \R^2 R2

  • Because the transformations are linear, they are determined completely by what they do to the columns of I 2 I_2 I2.

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EXAMPLE 3

Let T T T : : : R 2 → R 2 \mathbb R^2\rightarrow \mathbb R^2 R2R2 be the transformation that rotates each point in R 2 \mathbb R^2 R2 about the origin through an angle ϕ \phi ϕ, with counterclockwise rotation for a positive angle. We could show geometrically that such a transformation is linear. Find the standard matrix A A A of this transformation.

SOLUTION
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Existence and Uniqueness Questions

  • The concept of a linear transformation provides a new way to understand the existence and uniqueness questions asked earlier.

Existence question (满射)

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  • Does T T T map R n \mathbb R^n Rn onto R m \mathbb R^m Rm?” is an existence question.

Uniqueness question (单射)

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  • Is T T T one-to-one?” is a uniqueness question.

EXAMPLE 4

  • The projection transformations shown in Table 4 are not one-to-one and do not map R 2 \mathbb R^2 R2 onto R 2 \mathbb R^2 R2. The transformations in Tables 1, 2, and 3 are one-to-one and do map R 2 \mathbb R^2 R2 onto R 2 \mathbb R^2 R2.

EXAMPLE 5

Let T T T be the linear transformation whose standard matrix is
A = [ 1 − 4 8 1 0 2 − 1 3 0 0 0 5 ] A=\begin{bmatrix}1&-4&8&1\\0&2&-1&3\\0&0&0&5\end{bmatrix} A=100420810135Does T T T map R 4 \mathbb R^4 R4 onto R 3 \mathbb R^3 R3? Is T T T a one-to-one mapping?

SOLUTION

  • Since A A A happens to be in echelon form, we can see at once that A A A has a pivot position in each row. Therefore, T T T maps R 4 \mathbb R^4 R4 onto R 3 \mathbb R^3 R3.
  • However, since the equation A x = b A\boldsymbol x = \boldsymbol b Ax=b has a free variable, each b \boldsymbol b b is the image of more than one x \boldsymbol x x. That is, T T T is not one-to-one.

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PROOF

  • 必要性: Since T T T is linear, T ( 0 ) = 0 T(\boldsymbol 0) =\boldsymbol 0 T(0)=0. If T is one-to-one, then the equation T ( x ) = 0 T(\boldsymbol x) =\boldsymbol 0 T(x)=0 has at most one solution and hence only the trivial solution.
  • 充分性: If T T T is not one-to-one, then there is a b \boldsymbol b b that is the image of at least two different vectors in R n \mathbb R^n Rn—say, u \boldsymbol u u and v \boldsymbol v v. But then, since T T T is linear, T ( u − v ) = T ( u ) − T ( v ) = b − b = 0 T(\boldsymbol u-\boldsymbol v)=T(\boldsymbol u)-T(\boldsymbol v)=\boldsymbol b-\boldsymbol b=\boldsymbol 0 T(uv)=T(u)T(v)=bb=0. The vector u − v \boldsymbol u-\boldsymbol v uv is not zero. Hence the equation T ( x ) = 0 T(\boldsymbol x)=\boldsymbol 0 T(x)=0 has more than one solution.

也可以这样证明:设 A A A 为线性变换 T T T 对应的标准矩阵,则 T T T 为单射 ⇔ \Leftrightarrow A x = b A\boldsymbol x=\boldsymbol b Ax=b 有唯一解 ⇔ \Leftrightarrow A x = 0 A\boldsymbol x=\boldsymbol 0 Ax=0 只有平凡解 ⇔ \Leftrightarrow T ( x ) = 0 T(\boldsymbol x)=\boldsymbol 0 T(x)=0 只有平凡解


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可以发现,只有当 T T T 是方阵的时候 T T T 才可能是双射 ( T T T 的标准矩阵 A A A 的每行每列都必须要有主元)


EXAMPLE 6

Let T ( x 1 , x 2 ) = ( 3 x 1 + x 2 , 5 x 1 + 7 x 2 , x 1 + 3 x 2 ) T(x_1,x_2)=(3x_1+x_2,5x_1+7x_2,x_1+3x_2) T(x1,x2)=(3x1+x2,5x1+7x2,x1+3x2). Show that T T T is a one-to-one linear transformation. Does T T T map R 2 \mathbb R^2 R2 onto R 3 \mathbb R^3 R3?

SOLUTION

  • T T T is indeed a linear transformation, with its standard matrix A A A shown in (4).
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  • The columns of A A A are linearly independent because they are not multiples. By Theorem 12(b), T T T is one-to-one.
  • To decide if T T T is onto R 3 \mathbb R^3 R3, examine the span of the columns of A A A. Since A A A is 3 × 2 3 \times 2 3×2, the columns of A A A span R 3 \mathbb R^3 R3 if and only if A A A has 3 pivot positions. This is impossible, since A A A has only 2 columns. So the columns of A A A do not span R 3 \mathbb R^3 R3, and the associated linear transformation is not onto R 3 \mathbb R^3 R3.
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