91影视

To show that 饾湋 is linear, we need to prove that for any 饾憹, 饾憺 鈭� 饾憛[饾憢] and any scalars 饾浖, 饾浗, we have 饾湋(饾浖饾憹 + 饾浗饾憺) = 饾浖饾湋(饾憹) + 饾浗饾湋(饾憺). Let's begin: By the definition of 饾湋, $$ \varphi(\alpha p + \beta q) = (\alpha p + \beta q)(a) + (\alpha p + \beta q)' (a)(X-a) $$ We'll calculate both parts of the right-hand side separately. \((i)\: (\alpha p + \beta q)(a) = \alpha p(a) + \beta q(a)\) \((ii)\: (\alpha p + \beta q)' (a) = \alpha p'(a) + \beta q'(a)\) Therefore, combining (i) and (ii), $$ \varphi(\alpha p + \beta q) = \alpha p(a) + \beta q(a) + (\alpha p'(a) + \beta q'(a))(X-a) $$ Next, we compute 饾浖饾湋(饾憹) + 饾浗饾湋(饾憺) $$ \alpha \varphi(p) + \beta \varphi(q) = \alpha [p(a) + p'(a)(X-a)] + \beta [q(a) + q'(a)(X-a)] $$ $$ = \alpha p(a) + \beta q(a) + \alpha p'(a)(X-a) + \beta q'(a)(X-a) $$ Since both expressions are equal, we can conclude that 饾湋 is linear.

To find the representation matrix of 饾湋 with respect to the basis \(饾惛_{3} = (1, 饾憢, 饾憢^2, 饾憢^3)\), we need to determine 饾湋(饾憭_饾憱)\( for all elements \)饾憭_饾憱$ of the basis and express the result in terms of the basis coordinates. 1. For 饾憭_1 = 1: $ \varphi(1) = 1(a) + 1'(a)(X-a) = a $ 2. For 饾憭_2 = 饾憢: $ \varphi(X) = X(a) + X'(a)(X-a) = aX + 1(X-a) = 1 + (a-1)X $ 3. For 饾憭_3 = 饾憢^2: $ \varphi(X^2) = X^2(a) + X^2'(a)(X-a) = a^2 + 2a(X-a) = a^2 + 2aX - 2a^2 = (2a-2a^2) + 2aX $ 4. For 饾憭_4 = 饾憢^3: $ \varphi(X^3) = X^3(a) + X^3'(a)(X-a) = a^3 + 3a^2(X-a) = a^3 + 3a^2X - 3a^3 = (3a^2-3a^3) + 3a^2X $ Expressing the results as coordinates in \(饾惛_{3}\), we obtain: $ \varphi(1) = (a,0,0,0), \: \varphi(X) = (1,a-1,0,0), \: \varphi(X^2) = (2a-2a^2,2a,0,0), \: \varphi(X^3) = (3a^2-3a^3,3a^2,0,0) $ Thus, the representation matrix of 饾湋 with respect to the basis \(饾惛_{3}\) is: $$ \begin{pmatrix} a & 1 & 2a-2a^2 & 3a^2-3a^3 \\ 0 & a-1 & 2a & 3a^2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} $$

First, we diagonalize the matrix obtained in Step 2: $$ \begin{pmatrix} a & 1 & 2a-2a^2 & 3a^2-3a^3 \\ 0 & a-1 & 2a & 3a^2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & a-1 & 2a & 3a^2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} $$ This diagonal matrix has the eigenvalues \(1\), \(a-1\), \(0\), and \(0\). The associated eigenvectors are: 1. For 位 = 1: \((1, X-a, X^2(a), X^3(a^2))\) 2. For 位 = \(a-1\): \((0, 1, 2X, 3X^2)\) 3. For 位 = 0: \((0, 0, X^2, X^3)\) 4. For 位 = 0: \((0, 0, 0, X^3)\) Thus, a possible ordered basis \(饾惖\) of 饾憛[饾憢] for which the representation matrix of 饾湋 is diagonal is: $$ B = (1, X-a, X^2, X^3) $$