## 矩陣理論Homework 1

* Given $f \in P_n$, define $\phi_f^m : P_m \to P_{n+m}$ such that, $\forall g \in P_m$, $\phi_f^m(g) = f\cdot g$
then we have the matrix representation of $\phi_f^m$ ,$C_m(f)$ w.r.t. standard basis, and
$C_m(f) =
\begin{bmatrix}
a_0 \\
a_1 & a_0\\
\vdots & a_1 & \ddots & a_0 \\
a_n & \vdots & \ddots & a_1 \\
& a_n & \ddots & \vdots \\
& & & a_n
\end{bmatrix}$, for $f(x) = a_nx^n + \cdots +a_1x + a0$
and $C_m(f)$ is called the convolution matrix of $f$ in degree $m$
and $C_m(f)$ is $(n+m+1)\times(m+1)$
---
**Find the matrix representation of $\phi_f^m$ w.r.t. chebyshev basis**

* Let $f \in P_4$, $f(x) = a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$, let $\phi_f^2 : P_3 \to P_{4+3}$

then the matrix representation of $f$, $C_3(f) = \begin{bmatrix}
a_0\\
a_1 & a_0\\
a_2 & a_1 & a_0\\
a_3 & a_2 & a_1 & a_0\\
a_4 & a_3 & a_2 & a_1\\
& a_4 & a_3 & a_2\\
& & a_4 & a_3\\
& & & a_4\\
\end{bmatrix}$

* And the Chebyshev polynomial is :
$T_0 = 1$
$T_1 = x$
$T_2 = 2x^2$
$T_3 = 4x^3-3x$
$\vdots$
$T_n = 2xT_{n-1} - T_{n-2}$

*We have the transition matrix $S$ from Chebyshev basis(Ch) to standard basis(E) in order 3
$T_0 = 1 = 1\cdot 1 + 0\cdot x + 0\cdot x^2 + 0\cdot x^3$
$T_1 = x = 0\cdot 1 + 1\cdot x + 0\cdot x^2 + 0\cdot x^3$
$T_2 = 2x^2 = 0\cdot 1 + 0\cdot x + 2\cdot x^2 + 0\cdot x^3$
$T_3 = 4x^3 - 3x = 0\cdot 1 - 3\cdot x + 0\cdot x^2 + 4\cdot x^3$

then, $S =
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & -3 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 4 \\
\end{bmatrix}$

(in same method)
*We have the transition matrix $T$ from Chebyshev basis(Ch) to standard basis(E) in order 4+3
$T_4 = 8x^4 - 8x^2 + 1$
$T_5 = 16x^5 - 20x^3 + 5x$
$T_6 = 32x^6 - 48x^4 + 18x^2 - 1$
$T_7 = 64x^7 - 112x^5 + 56x^3 - 7x$

then, $T =
\begin{bmatrix}
1 & 0 & 0 & 0 & 1 & 0 & -1 & 0 \\
0 & 1 & 0 & -3 & 0 & 5 & 0 & -7 \\
0 & 0 & 2 & 0 & -8 & 0 & 18 & 0 \\
0 & 0 & 0 & 4 & 0 & -20 & 0 & 56 \\
0 & 0 & 0 & 0 & 8 & 0 & -48 & 0 \\
0 & 0 & 0 & 0 & 0 & 16 & 0 & -112\\
0 & 0 & 0 & 0 & 0 & 0 & 32 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 64 \\
\end{bmatrix}$

and the transition matrix from standard basis(E) to hebyshev basis(Ch) is $T^{-1}$
$T^{-1} =
\begin{bmatrix}
1 & 0 & 0 & 0 & \frac{-1}{8} & 0 & \frac{-5}{32} & 0 \\
0 & 1 & 0 & \frac{3}{4} & 0 & \frac{5}{8} & 0 & \frac{35}{64} \\
0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & \frac{15}{32} & 0 \\
0 & 0 & 0 & \frac{1}{4} & 0 & \frac{5}{16} & 0 & \frac{21}{64} \\
0 & 0 & 0 & 0 & \frac{1}{8} & 0 & \frac{3}{16} & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{1}{16} & 0 & \frac{7}{64} \\
0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{32} & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{64}
\end{bmatrix}$