Exercise MR.C20

Compute the matrix representation of \(T\) relative to the bases \(B\) and \(C\).

\[T:P_3\rightarrow \mathbb{C} ^3,\quad T(a+bx+cx^2+dx^3)= \begin{bmatrix} 2a & -3b & +4c & -2d \\ 1a & 1b & -1c & 1d \\ 3a & 0b & 2c & -3d \end{bmatrix}\] \[B = \{1,x,x^2,x^3\}\quad C= \left\{ \begin {bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin {bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin {bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \right\} \]

C is upper triangle, so it is easy to solve the systems of equations to find each column in the matrix representation

Col 1
\[\begin {bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} = \left\{ X_1 \centerdot \begin {bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, X_2 \centerdot \begin {bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, X_3 \centerdot \begin {bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \right\} \] working from the bottom up:

  1. \(X_3 \centerdot 1 = 3,\quad X_3 = 3\)
  2. \(X_2 \centerdot 1 + 3 = 1,\quad X_2 = -2\)
  3. \(X_1 \centerdot 1 -2 + 3 = 2,\quad X_1 = 1\)

this gives the vector: \[\begin {bmatrix} 1 \\ -2 \\ 3 \end{bmatrix}\]

Col 2 \[\begin {bmatrix} -3 \\ 1 \\ 0 \end{bmatrix} = \left\{ X_1 \centerdot \begin {bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, X_2 \centerdot \begin {bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, X_3 \centerdot \begin {bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \right\} \]

working from the bottom up:

  1. \(X_3 \centerdot 1 = 0,\quad X_3 = 0\)
  2. \(X_2 \centerdot 1 + 0 = 1,\quad X_2 = 1\)
  3. \(X_1 \centerdot 1 + 1 + 0 = -3,\quad X_1 = -4\)

this gives the vector: \[\begin {bmatrix} -4 \\ 1 \\ 0 \end{bmatrix}\]

Col 3 \[\begin {bmatrix} 4 \\ -1 \\ 2 \end{bmatrix} = \left\{ X_1 \centerdot \begin {bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, X_2 \centerdot \begin {bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, X_3 \centerdot \begin {bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \right\} \]

working from the bottom up:

  1. \(X_3 \centerdot 1 = 2,\quad X_3 = 2\)
  2. \(X_2 \centerdot 1 + 2 = -1,\quad X_2 = -3\)
  3. \(X_1 \centerdot 1 - 3 + 2 = 4,\quad X_1 = 5\)

this gives the vector: \[\begin {bmatrix} 5 \\ -3 \\ 2 \end{bmatrix}\]

Col 4 \[\begin {bmatrix} -2 \\ 1 \\ -3 \end{bmatrix} = \left\{ X_1 \centerdot \begin {bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, X_2 \centerdot \begin {bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, X_3 \centerdot \begin {bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \right\} \]

working from the bottom up:

  1. \(X_3 \centerdot 1 = -3,\quad X_3 = -3\)
  2. \(X_2 \centerdot 1 -3 = 1,\quad X_2 = 4\)
  3. \(X_1 \centerdot 1 - 3 + 4 = -2,\quad X_1 = -3\)

this gives the vector: \[\begin {bmatrix} -3 \\ 4 \\ -3 \end{bmatrix}\]

Together, these 4 vectors build the new matrix representation:

\[M_{B,C}^{T} = \begin {bmatrix} 1 & -4 & 5 & -3 \\ -2 & 1 & -3 & 4 \\ 3 & 0 & 2 & -3 \end{bmatrix} \]