In each of the following cases, let \(T\) be the linear operator on \(\RR^2\) which is represented by the matrix \(A\) in the standard ordered basis for \(\RR^2\), and let \(U\) be the linear operator on \(C^2\) represented by \(A\) in the standard ordered basis. Find the characteristic polynomial for \(T\) and that for \(U\), find the characteristic values of each operator, and for each such characteristic value \(c\) find a basis for the corresponding space of characteristic vectors. \[ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \quad A = \begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}. \]
Solution
For \[ A_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}, \] we have \[ \chi(x)=\det(xI-A_1)=x(x-1). \] Thus both over \(\mathbb R\) and \(\mathbb C\) the characteristic values are \(0,1\), with eigenspaces \[ E_0=\operatorname{span}\{(0,1)\},\qquad E_1=\operatorname{span}\{(1,0)\}. \]
For \[ A_2=\begin{bmatrix}2&3\\-1&1\end{bmatrix}, \] \[ \chi(x)=\det\begin{bmatrix}x-2&-3\\1&x-1\end{bmatrix}=x^2-3x+5. \] Its roots are \[ \lambda_\pm=\frac{3\pm i\sqrt{11}}2. \] Hence \(T\) on \(\mathbb R^2\) has no characteristic values and therefore no nonzero characteristic-vector spaces. For \(U\) on \(\mathbb C^2\), solving \((A_2-\lambda I)v=0\) gives, for example, \[ E_{\lambda_+}=\operatorname{span}_{\mathbb C}\{(-1-i\sqrt{11},\,2)\}, \qquad E_{\lambda_-}=\operatorname{span}_{\mathbb C}\{(-1+i\sqrt{11},\,2)\}. \]
For \[ A_3=\begin{bmatrix}1&1\\1&1\end{bmatrix}, \] \[ \chi(x)=\det\begin{bmatrix}x-1&-1\\-1&x-1\end{bmatrix}=x(x-2). \] Thus over both fields the characteristic values are \(0,2\), with \[ E_0=\operatorname{span}\{(-1,1)\},\qquad E_2=\operatorname{span}\{(1,1)\}. \]