ns2675 said:
Is the matrix M^C_C(phi) supposed to be the matrix of phi in the ordered basis C? If so, your matrix is incorrect; 1+x, 1-x, and 1+x^2 are eigenvectors of phi, with eigenvalues 1, 3, and 2, respectively. The matrix you are looking for is a diagonal matrix.
Your approach for (iii) is correct; simply multiply the matrix of phi in the basis C by the coordinate matrix v; that will give you the coordinate matrix of phi(v).
Well this is my approach to the whole question:
Part (i)
\phi (1) = 2-x = 2(1) -1(x) + 0(x^2)
\phi (x) = -1 + 2x = -1(1) + 2(x) + 0(x^2)
\phi (x^2) = x + 2x^2 = 0(1) + 1(x) + 2(x^2)
M^{\mathcal B}_{\mathcal B}(\phi) = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \\ \end{bmatrix}
Part (ii)
Since {\mathcal C} has size 3 and \text{dim}(\mathbb{R}_2[x])=3 it is enough to show that \mathcal C} is either linearly independent or spans V . For completeness let's do both.
{\mathcal C} is linearly independent if
\alpha (1+x) + \beta (1-x) + \gamma (1+x^2) = 0 \Rightarrow \alpha = \beta = \gamma = 0
\alpha + \alpha x +\beta - \beta x + \gamma + \gamma x^2 = 0
\alpha + \beta + \gamma + (\alpha - \beta)x + \gamma x^2 = 0
Equate coefficients.
\alpha + \beta + \gamma = 0 (1)
\alpha - \beta = 0 \Rightarrow \alpha = \beta (2)
\gamma = 0 (3)
Subbing (2) and (3) in (1) gives 2\alpha = 0 \Rightarrow \alpha = 0 \Rightarrow \beta = 0 from (2).
Hence \alpha = \beta = \gamma = 0 and \mathcal C is linearly independent.
\mathcal C spans V if
a_1 (1+x) + a_2 (1-x) + a_3 (1+x^2) = b_1 + b_2 x + b_3 x^2
has at least one solution for every set of coefficients b_1,b_2,b_3
a_1 + a_2 + a_3 + (a_1 - a_2)x + a_3x^2 = b_1 + b_2 x + b_3 x^2
Equating coefficients yields the following system of equations (*):
a_1 + a_2 + a_3 = b_1
a_1 - a_2 = b_2
a_3 = b_3
Transforming the augmented matrix
\begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ \end{bmatrix}
into row echelon form we get
\begin{bmatrix} 1 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & 1 & 0 & 0 & 1 \\ \end{bmatrix}
which corresponds to the following system of equations:
a_1 = \frac{1}{2}b_1 + \frac{1}{2}b_2 -\frac{1}{2}b_3
a_2 = \frac{1}{2}b_1 -\frac{1}{2}b_2 -\frac{1}{2}b_3
a_3 = b_3
This system {and therefore the system (*)} are consistent for all RHS values, therefore \mathcal C spans V .
We can therefore conclude that \mathcal C is certainly a basis for V .
\phi (1+x) = (2-1) + (-1+2)x = 1 + x = 1(1) + 1(x) + 0(x^2)
\phi (1-x) = (2+1) + (-1-2)x = 3-3x = 3(1) - 3(x) + 0(x^2)
\phi (1+x^2) = 2 + (-1+1)x + 2x^2 = 2+2x^2 = 2(1) + 0(x) + 2(x^2)
M^{\mathcal C}_{\mathcal C} (\phi) = \begin{bmatrix} 1 & 3 & 2 \\ 1 & -3 & 0 \\ 0 & 0 & 2 \end{bmatrix}
Part (iii)
M^{\mathcal C}_{\mathcal C} (\phi) {\bf v} = {\bf w} where {\bf w} is the coordinate vector of \phi (v) with respect to \mathcal C and {\bf v} is the coordinate vector of v\in V .
{\bf v} = \begin{bmatrix} 1 \\ -3 \\ \pi \end{bmatrix}
So {\bf w} = \begin{bmatrix} 1 & 3 & 2 \\ 1 & -3 & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ -3 \\ \pi \end{bmatrix} = \begin{bmatrix} 2\pi -8 \\ 10 \\ 2\pi \end{bmatrix}