12.13 Change basis

Let \(T: \mathbb{R}^3 \to \mathbb{R}^3\), \(T(x, y, z) = (x + y, x - 2y + z, -y + 2z)\) and consider the basis \(B = \{ (1, 0, 1), (0, 1, -2), (-1, -1, 0) \}\) of \(\mathbb{R}^3\). Find \([T]_B\).
One way: \([T]_B =\begin{pmatrix}[T(1, 0, 1)]_B &[T(0, 1, -2)]_B &[T(-1, -1, 0)]_B\end{pmatrix}\)

Another way: Find \([T]_B\) for \(C = \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}\) the canonical basis of \(\mathbb{R}^3\).

Then we will have \([T]_B = (P_B^C)^{-1} [T]_C P_B^C\)

\([T]_C = \begin{pmatrix}[T(1, 0, 0)]_C & [T(0, 1, 0)]_C & [T(0, 0, 1)]_C\end{pmatrix}=\begin{pmatrix}T(1,0,0)^{T} & T(0,1,0)^{T} & T(0,0,1)^{T}\end{pmatrix}=\begin{pmatrix}1 & 1 & 0\\ 1 & -2 & 1\\ 0 & -3 & 2\end{pmatrix}\)

\(P_B^C = \begin{pmatrix}[(1, 0, 1)]_C & [(0, 1, -2)]_C & [(-1, -1, 0)]_C\end{pmatrix}= \begin{pmatrix}1 & 0 & -1 \\0 & 1 & -1 \\1 & -2 & 0\end{pmatrix}\)and \((P_B^C)^{-1} = \begin{pmatrix}2 & -2 & -1 \\1 & -1 & -1 \\1 & -2 & -1\end{pmatrix}\)

Therefore \([T]_B = (P_B^C)^{-1} [T]_C P_B^C = \begin{pmatrix}-4 & 17 & -9 \\-3 & 12 & -6 \\5 & 16 & -7\end{pmatrix}\)
Let \(T: \mathbb{R}^2 \to \mathbb{R}^2\), \(T(x, y) = (x + 2y, 3x - y)\) and consider the bases \(B = \{ (1, 1), (1, 0) \}\) and \(B' = \{ (4, 7), (4, 8) \}\) of \(\mathbb{R}^2\).
1. Find \([T]_B\).
  
  In the same way as exercise (1):
  \([T]_B = (P_B^C)^{-1} [T]_C P_B^C\)\([T]_C =\begin{pmatrix}T(1, 0) & T(0, 1)\end{pmatrix}= \begin{pmatrix}1 & 2 \\3 & -1\end{pmatrix}\)
  
  \(P_B^C =\begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix}\)and \((P_B^C)^{-1} =\begin{pmatrix}0 & -1 \\-1 & 1\end{pmatrix} =\begin{pmatrix}0 & 1 \\1 & -1\end{pmatrix}\)
  
  So \([T]_B = (P_B^C)^{-1} [T]_C P_B^C =\begin{pmatrix}2 & 3 \\1 & -2\end{pmatrix}\)
  
  Let's verify this is correct:
  
  \([T]_B\begin{pmatrix}1 \\1\end{pmatrix}_B = [T(1, 1)]_B\)\(\begin{pmatrix}2 & 3 \\1 & -2\end{pmatrix}\begin{pmatrix}1 \\1\end{pmatrix}=\begin{pmatrix}3 \\2\end{pmatrix}_B\)
  
  This corresponds to \(2(1, 1) + 1(1, 0) = (3, 2)\)This verifies the correctness.
  
  Exercise \([T]_B [(1, 0)]_B = [T((1, 0))]_B\) 2. Write \([T]_{B'} = P^{-1} [T]_B P\) for \(P\), an invertible matrix.
  
  One way: Find \(P=P_{B^{\prime}}^{B}\) (more complicated).
  
  Another way: In the exact same way as (a) \([T]_{B'} = (P_{B'}^C)^{-1} [T]_C P_{B'}^C (\star1)\)
  
  By (a), we have \([T]_{B}=(P_{B}^{C})^{-1}[T]_{C}P_{B}^{C}\), then \([T]_{C}=P_{B}^{C}[T]_{B}(P_{B}^{C})^{-1}(\star2)\)
  
  By \((\star1),(\star2)\) we have that \([T]_{B^{\prime}}=(P_{B^{\prime}}^{C})^{-1}P_{B}^{C}[T]_{B}(P_{B}^{C})^{-1}P_{B^{\prime}}^{C}\)
  
  So \([T]_{B'} = P^{-1} [T]_B P\) where \(P = (P_B^C)^{-1} P_{B'}^C\)
  
  The only thing left to find is \(P_{B'}^C\)
  
  \(P_{B'}^C = \begin{pmatrix} [ (4, 7) ]_C & [ (4, 8) ]_C \end{pmatrix}= \begin{pmatrix} 4 & 4 \\ 7 & 8 \end{pmatrix}\)
  
  So \(P = \begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 4 & 4 \\ 7 & 8 \end{pmatrix}= \begin{pmatrix} 7 & 8 \\ -3 & -4 \end{pmatrix}\)

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