Let \(T\) be the linear operator on \(\RR^3\) defined by \[ T(x_1, x_2, x_3) = (3x_1 + x_3,\; -2x_1 + x_2,\; -x_1 + 2x_2 + 4x_3). \]
What is the matrix of \(T\) in the standard ordered basis for \(\RR^3\)?
What is the matrix of \(T\) in the ordered basis \[ \{\alpha_1, \alpha_2, \alpha_3\} \] where \(\alpha_1 = (1, 0, 1)\), \(\alpha_2 = (-1, 2, 1)\), and \(\alpha_3 = (2, 1, 1)\)?
Prove that \(T\) is invertible and give a rule for \(T^{-1}\) like the one which defines \(T\).
Solution
In the standard basis, \[ A=[T]=\begin{bmatrix}3&0&1\\-2&1&0\\-1&2&4\end{bmatrix}. \] Let \[ P=\begin{bmatrix}1&-1&2\\0&2&1\\1&1&1\end{bmatrix} \] be the matrix whose columns are the given basis vectors. Then \[ [T]_{\mathfrak B}=P^{-1}AP =\begin{bmatrix} \frac{17}{4}&\frac{35}{4}&\frac{11}{2}\\ -\frac34&\frac{15}{4}&-\frac32\\ -\frac12&-\frac72&0 \end{bmatrix}. \]
Since \(\det A=9\ne0\), \(T\) is invertible. Solving \(T(x_1,x_2,x_3)=(y_1,y_2,y_3)\), equivalently using \[ A^{-1}=\begin{bmatrix} \frac49&\frac29&-\frac19\\ \frac89&\frac{13}{9}&-\frac29\\ -\frac13&-\frac23&\frac13 \end{bmatrix}, \] gives \[ T^{-1}(y_1,y_2,y_3)= \left( \frac{4y_1+2y_2-y_3}{9}, \frac{8y_1+13y_2-2y_3}{9}, \frac{-y_1-2y_2+y_3}{3} \right). \]