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6.5 Matrix expression of linear transformation

Define 6

Let TTT be a linear transformation in the linear space VnV_nVn, in which a basis α1,α2… , alpha n \ alpha_1 \ alpha_2,… , \ alpha_n alpha 1, alpha 2,… ,αn, if the image of this basis under transformation TTT (expressed linearly by this basis) is:


{ T ( Alpha. 1 ) = a 11 Alpha. 1 + a 21 Alpha. 2 + . . . + a n 1 Alpha. n T ( Alpha. 2 ) = a 12 Alpha. 1 + a 22 Alpha. 2 + . . . + a n 2 Alpha. n . . . . . . . . . T ( Alpha. n ) = a 1 n Alpha. 1 + a 2 n Alpha. 2 + . . . + a n n Alpha. n (1) \begin{cases} T(\alpha_1)=a_{11}\alpha_1+a_{21}\alpha_2+… +a_{n1}\alpha_n\\ T(\alpha_2)=a_{12}\alpha_1+a_{22}\alpha_2+… +a_{n2}\alpha_n\\ ……… \\ T(\alpha_n)=a_{1n}\alpha_1+a_{2n}\alpha_2+… +a_{nn}\alpha_n\\ \end{cases} \tag{1}

T remember (alpha 1, alpha 2,… , alpha n) = (T (alpha 1), T (alpha 2),… T (n), T (alpha) \ alpha_1 \ alpha_2,… ,\alpha_n)=(T(\alpha_1),T(\alpha_2),… T, T (\ alpha_n)) (alpha 1, alpha 2,… , alpha n) = (T (alpha 1), T (alpha 2),… ,T(αn)), then equation (1) can be expressed as


T ( Alpha. 1 . Alpha. 2 . . . . . Alpha. n ) = ( Alpha. 1 . Alpha. 2 . . . . . Alpha. n ) A T(\alpha_1,\alpha_2,… ,\alpha_n)=(\alpha_1,\alpha_2,… ,\alpha_n)A

Among them


A = [ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . a n 1 a n 2 . . . a n n ] A=\begin{bmatrix} a_{11} & a_{12} &… & a_{1n}\\ a_{21} & a_{22} & … &a_{2n}\\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{bmatrix}

So, AAA is called the linear transformation TTT at the basis α1,α2… , alpha n \ alpha_1 \ alpha_2,… , \ alpha_n alpha 1, alpha 2,… Alpha n matrix

Theorem 2

Let’s take two bases in the linear space VnV_nVn


Alpha. 1 . Alpha. 2 . . . . . Alpha. n , Beta. 1 . Beta. 2 . . . . . Beta. n \alpha_1,\alpha_2,… , \ alpha_n, \ beta_1 \ beta_2,… ,\beta_n

By the base alpha 1, alpha 2,… , alpha n \ alpha_1 \ alpha_2,… , \ alpha_n alpha 1, alpha 2,… , alpha n to the beta 1, beta 2,… , beta n \ beta_1 \ beta_2,… , \ beta_n beta 1, beta 2,… B=P−1APB=P^{-1}APB=P−1AP

prove

Alpha 1, alpha 2,… , alpha n \ alpha_1 \ alpha_2,… , \ alpha_n alpha 1, alpha 2,… , alpha n to the beta 1, beta 2,… , beta n \ beta_1 \ beta_2,… , \ beta_n beta 1, beta 2,… , the transition matrix of βn is PPP, and


( Beta. 1 . Beta. 2 . . . . . . Beta. n ) = ( Alpha. 1 . Alpha. 2 . . . . . Alpha. n ) P . P reversible (\beta_1,\beta_2,…. ,\beta_n)=(\alpha_1,\alpha_2,… ,\alpha_n)P, P reversible

The deformation is


( Alpha. 1 . Alpha. 2 . . . . . Alpha. n ) = ( Beta. 1 . Beta. 2 . . . . . . Beta. n ) P – 1 (\alpha_1,\alpha_2,… ,\alpha_n)=(\beta_1,\beta_2,…. ,\beta_n)P^{-1}

And the matrix of the linear transformation TTT in VnV_nVn under these two bases is A, BA, BA, B, and


T ( Alpha. 1 . Alpha. 2 . . . . . Alpha. n ) = ( Alpha. 1 . Alpha. 2 . . . . . Alpha. n ) A T ( Beta. 1 . Beta. 2 . . . . . Beta. n ) = ( Beta. 1 . Beta. 2 . . . . . Beta. n ) B T(\alpha_1,\alpha_2,… ,\alpha_n)=(\alpha_1,\alpha_2,… ,\alpha_n)A\\ \quad \\ T(\beta_1,\beta_2,… ,\beta_n)=(\beta_1,\beta_2,… ,\beta_n)B

So, there are


( Beta. 1 . Beta. 2 . . . . . Beta. n ) B = T ( Beta. 1 . Beta. 2 . . . . . Beta. n ) = T ( ( Alpha. 1 . Alpha. 2 . . . . . Alpha. n ) P ) = T ( Alpha. 1 . Alpha. 2 . . . . . Alpha. n ) P = ( Alpha. 1 . Alpha. 2 . . . . . Alpha. n ) A P = ( Beta. 1 . Beta. 2 . . . . . . Beta. n ) P – 1 A P \quad(\beta_1,\beta_2,… ,\beta_n)B=T(\beta_1,\beta_2,… ,\beta_n)=T((\alpha_1,\alpha_2,… ,\alpha_n)P)\\ \quad \\ =T(\alpha_1,\alpha_2,… ,\alpha_n)P=(\alpha_1,\alpha_2,… ,\alpha_n)AP\\ \quad \\ =(\beta_1,\beta_2,…. ,\beta_n)P^{-1}AP

Because (beta 1, beta 2,… , beta, n) (\ beta_1, \ beta_2,… , \ beta_n) (beta 1, beta 2,… Beta n is linearly independent, so


B = P – 1 A P B=P^{-1}AP

Beta = (beta 1, beta 2,… N) and beta \ beta = (\ beta_1 \ beta_2,… , \ beta_n) beta = (beta 1, beta 2,… , beta n) linear independence, can explain the beta \ beta beta reversible, is beta – 1 \ beta ^ {1} beta – 1 beta B = beta P – 1 in equation AP \ beta = B \ beta P ^ {1} AP beta B = beta – 1 P AP, At the same time left by beta – 1 \ beta ^ {1} beta – 1 can be B = P – 1 APB = P ^ {1} APB = P – 1 ap

Definition of 7

The dimension of the image space T(Vn)T(V_n)T(Vn) of the linear space TTT is called the rank of the linear transformation TTT

  • If AAA is the matrix of TTT, then the rank of TTT is R(A)R(A)R(A).
  • If the rank of TTT is RRR, then the dimension of TTT’s kernel STS_TST is N − Rn-Rn − R

For example,

Example 11

In P[x]3P[x] 3P[x] 3P[x]3, take the basis


p 1 = x 3 . p 2 = x 2 . p 3 = x . p 4 = 1 p_1=x^3,p_2=x^2,p_3=x,p_4=1

Find the matrix of the differential operation DDD

Answer:

Apply a linear transformation to each of these bases (DDD in this case, the first derivative)


{ D ( p 1 ) = D ( x 3 ) = ( x 3 ) ‘ = 3 x 2 = 0 p 1 + 3 p 2 + 0 p 3 + 0 p 4 D ( p 2 ) = 2 x = 0 p 1 + 0 p 2 + 2 p 3 + 0 p 4 D ( p 3 ) = 1 = 0 p 1 + 0 p 2 + 0 p 3 + 1 p 4 D ( p 4 ) = 0 = 0 p 1 + 0 p 2 + 0 p 3 + 0 p 4 \begin{cases} D(p_1)=D(x^3)=(x^3)^{‘}=3x^2=0p_1+3p_2+0p_3+0p_4\\ \quad\\ D(p_2)=2x=0p_1+0p_2+2p_3+0p_4\\ \quad\\ D(p_3)=1=0p_1+0p_2+0p_3+1p_4\\ \quad\\ D(p_4)=0=0p_1+0p_2+0p_3+0p_4\\ \end{cases}

Therefore,


D ( p 1 . p 2 . p 3 . p 4 ) = ( p 1 . p 2 . p 3 . p 4 ) [ 0 0 0 0 3 0 0 0 0 2 0 0 0 0 1 0 ] D(p_1,p_2,p_3,p_4)=(p_1,p_2,p_3,p_4)\begin{bmatrix} 0 & 0 & 0 & 0\\ 3 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 1 & 0\\ \end{bmatrix}

So the matrix of the differential operation DDD under this basis is zero


A = [ 0 0 0 0 3 0 0 0 0 2 0 0 0 0 1 0 ] A=\begin{bmatrix} 0 & 0 & 0 & 0\\ 3 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 1 & 0\\ \end{bmatrix}

I take a linear transformation of PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI PI

conclusion

Description:

  • Refer to textbook “linear algebra” fifth edition tongji University mathematics department
  • With the book concept explanation combined with some of their own understanding and thinking

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