第 05 篇、矩陣觀點下的線性變換

$T:\begin{array}{rcl} R^2 & \longrightarrow & R^3 \\ (a_1,a_2) & \mapsto & (a_1+3a_2,0,2a_1-4a_2) \end{array}$
$\beta = \{ e_1,e_2 \}, \gamma = \{ e_1,e_2,e_3 \}$
$\left\{ \begin{array}{cl} T(1,0) & = (1,0,2) &= 1e_1+0e_2+2e_3 \\ T(0,1) & = (3,0,-4) &= 3e_1+0e_2-4e_3 \end{array} \right \} \Rightarrow \left[ T \right]_{\beta}^{\gamma} = \begin{bmatrix} 1 & 3 \\ 0 & 0 \\ 2 & -4 \end{bmatrix}$

$T:\begin{array}{rcl} P_2(R) & \longrightarrow & P_1(R) \\ f(x) & \mapsto & f'(x) \end{array}$
$\beta = \{ 1,x,x^2 \}, \gamma = \{ 1,x \}$
$\left\{ \begin{array}{cl} T(1) & = 0 &= 0 \times 1 + 0 x \\ T(x) & = 1 &= 1 \times 1 + 0 x \\ T(x^2) & = 2x &= 0 \times1 + 2 x \end{array} \right \} \Rightarrow \left[ T \right]_{\beta}^{\gamma} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix}$

令$T:\begin{array}{rcl} P_1(R) & \longrightarrow & P_2(R) \\ f(x) & \mapsto & \int_{0}^{x}f(t)dt \end{array}$
$U:\begin{array}{rcl} P_2(R) & \longrightarrow & P_1(R) \\ f(x) & \mapsto & f'(x) \end{array}$
$\alpha = \{ 1,x \}, \beta = \{ 1,x,x^2 \}, \gamma =\{ 1,x \}$
$(U \circ T)(ax+b) = ax+b, \forall ax+b \in P_1(R) \Rightarrow \left[ U \circ T \right]_{\alpha}^{\gamma} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

令$T:\begin{array}{rcl} M_2(R) & \longrightarrow & P_2(R) \\ \begin{pmatrix} a & b \\ c & d \end{pmatrix} & \mapsto & (a+b)+2dx+bx^2 \end{array}$
$U:\begin{array}{rcl} P_2(R) & \longrightarrow & M_2(R) \\ f(x) & \mapsto & \begin{pmatrix} f(0) & f(1) \\ f(2) & f(0) \end{pmatrix} \end{array}$
$(U \circ T)(\begin{pmatrix} a & b \\ c & d \end{pmatrix}) = U(a+b+2dx+bx^2) = \begin{pmatrix} a+b & a+2b+2d \\ a+5b+4d & a+b \end{pmatrix}$
$\Rightarrow \left[ U \circ T \right]_{\alpha}^{\gamma} = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & 2 & 0 & 2 \\ 1 & 5 & 0 & 4 \\ 1 & 1 & 0 & 0 \end{bmatrix}$

$V = M_2(R),\beta = \{ 1,x,x^2 \} , f(x) = 4+6x-7x^2 \Rightarrow \left[ f \right]_{\beta} = \begin{pmatrix} 4 \\ 6 \\ -7 \end{pmatrix}$
$\gamma = \{ 1,x-1,(x-1)^2 \} \Rightarrow \left[ f \right]_{\gamma} = \begin{pmatrix} 3 \\ -8 \\ -7 \end{pmatrix}$

$V = M_2(R),\beta = \{ E_{11},E_{12},E_{21}E_{22} \}$
若$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \in V \Rightarrow \left[ A \right]_{\gamma} = \begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix}$

$T:\begin{array}{rcl} P_2(R) & \longrightarrow & M_2(R) \\ f(x) & \mapsto & \begin{pmatrix} f(1)-f(2) & 0 \\ 0 & f(0) \end{pmatrix} \end{array} \Rightarrow \left[ T \right]_{\beta}^{\gamma} = \begin{bmatrix} 0 & -1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}$
若$f(x) = 1-x+x^2 \Rightarrow \left[ f \right]_{\beta} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, T(f) = \begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix} \Rightarrow \left[ T(f) \right]_{\gamma} = \begin{pmatrix} -2 \\ 0 \\ 0 \\ 1 \end{pmatrix}$
$\left[ T(f) \right]_{\gamma} = \left[ T \right]_{\beta}^{\gamma} \left[ f \right]_{\beta} \Rightarrow \begin{pmatrix} -2 \\ 0 \\ 0 \\ 1 \end{pmatrix} = \begin{bmatrix} 0 & -1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$

$T:\begin{array}{rcl} P_2(R) & \longrightarrow & P_1(R) \\ f(x) & \mapsto & f'(x) \end{array} \Rightarrow \left[ T \right]_{\beta}^{\gamma} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
若$f(x) = 2-4x+x^2 \Rightarrow \left[ f \right]_{\beta} = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}, T(f) = -4+2x \Rightarrow \left[ T(f) \right]_{\gamma} = \begin{pmatrix} -4 \\ 2 \end{pmatrix}$
$\left[ T(f) \right]_{\gamma} = \left[ T \right]_{\beta}^{\gamma} \left[ f \right]_{\beta} \Rightarrow \begin{pmatrix} -4 \\ 2 \end{pmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}\begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}$

$f(x) = x$ 的反變換$\Rightarrow f^{-1}(x) = x$
$f(x) = 0$ 的反變換$\Rightarrow$ 不存在

$f(x) = x+2$ 的反變換$\Rightarrow f^{-1}(x) = x-2$
$f(x) = 3x$ 的反變換$\Rightarrow f^{-1}(x) = \frac{1}{3}x$

$f(x) = x^2$ 的反變換$\Rightarrow$ 需討論一下
定義域$f: R_{ x\ge 0} \longrightarrow R_{ x\ge 0} \Rightarrow f^{-1}(x) = \sqrt{x}$

$f(x) = logx$ 的反變換$\Rightarrow f^{-1}(x) = 10^x$
$f(x) = lnx$ 的反變換$\Rightarrow f^{-1}(x) = e^x$

$f(x) = sinx$ 的反變換$\Rightarrow f^{-1}(x) = sin^{-1}x$

$T:\begin{array}{rcl} P_n(R) & \longrightarrow & P_{n-1}(R) \\ f(x) & \mapsto & f'(x) \end{array} U:\begin{array}{rcl} P_{n-1}(R) & \longrightarrow & P_n(R) \\ f(x) & \mapsto & \int^{x}_{0} f(t)dt \end{array}$
=> T、U 不互為反變換

$T:\begin{array}{rcl} P_1(R) & \longrightarrow & R^2 \\ a+bx & \mapsto & (a,a+b) \end{array} U:\begin{array}{rcl} R^2 & \longrightarrow & P_1(R) \\ (c,d) & \mapsto & c+(d-c)x \end{array}$
$(T \circ U)(c,d) = T(c+(d-c)x) = (c,d)$
$(U \circ T)(a+bx) = U(a,a+b) = a+bx$
$\left[ T \right]_{\beta}^{\gamma} = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} , \left[ U \right]_{\gamma}^{\beta} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$

$(\left[ T \right]_{\beta}^{\gamma})^{-1} =\left[ U \right]_{\gamma}^{\beta} \Leftrightarrow (\left[ U \right]_{\gamma}^{\beta})^{-1} = \left[ T \right]_{\beta}^{\gamma}$

$\left[ T \right]_{\beta} = R_{ \theta } = \begin{pmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{pmatrix}$
$\Rightarrow \left[ T^{-1} \right]_{\beta} = (\left[ T \right]_{\beta})^{-1} = \begin{pmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{pmatrix}$

$\left[ T \right]_{\beta} = M_{ \theta } = \begin{pmatrix} cos2\theta & sin2\theta \\ sin2\theta & -cos2\theta \end{pmatrix}$
$\Rightarrow \left[ T^{-1} \right]_{\beta} = (\left[ T \right]_{\beta})^{-1} = M_{\theta}$