Let \(T\) be the linear operator on \(C^2\) defined by \(T(x_1, x_2) = (x_1, 0)\). Let \(\mathfrak{B}\) be the standard ordered basis for \(C^2\) and let \(\mathfrak{B}' = \{\alpha_1, \alpha_2\}\) be the ordered basis defined by \(\alpha_1 = (1, i)\), \(\alpha_2 = (-i, 2)\).
What is the matrix of \(T\) relative to the pair \(\mathfrak{B}\), \(\mathfrak{B}'\)?
What is the matrix of \(T\) relative to the pair \(\mathfrak{B}'\), \(\mathfrak{B}\)?
What is the matrix of \(T\) in the ordered basis \(\mathfrak{B}'\)?
What is the matrix of \(T\) in the ordered basis \(\{\alpha_2, \alpha_1\}\)?
Solution
Let \[ A=\begin{bmatrix}1&0\\0&0\end{bmatrix} \] be the matrix of \(T\) in the standard basis, and let \[ P=\begin{bmatrix}1&-i\\ i&2\end{bmatrix} \] be the matrix whose columns are \(\alpha_1,\alpha_2\). Since \(\det P=1\), \(P\) is invertible.
For (a), the domain basis is standard and the codomain basis is \(\mathfrak B'\), so \[ [T]_{\mathfrak B,\mathfrak B'}=P^{-1}A =\begin{bmatrix}2&0\\-i&0\end{bmatrix}. \] For (b), \[ [T]_{\mathfrak B',\mathfrak B}=AP =\begin{bmatrix}1&-i\\0&0\end{bmatrix}. \] For (c), \[ [T]_{\mathfrak B'}=P^{-1}AP =\begin{bmatrix}2&-2i\\-i&-1\end{bmatrix}. \] If \(\mathfrak C=\{\alpha_2,\alpha_1\}\), then using the change-of-basis matrix obtained by swapping the columns of \(P\) gives \[ [T]_{\mathfrak C} =\begin{bmatrix}-1&-i\\-2i&2\end{bmatrix}. \]