Let \(V\) be the vector space of all \(2 \times 2\) matrices over the field of real numbers, and let \[ B = \begin{bmatrix} 2 & -2 \\ -1 & 1 \end{bmatrix}. \] Let \(W\) be the subspace of \(V\) consisting of all \(A\) such that \(AB = 0\). Let \(f\) be a linear functional on \(V\) which is in the annihilator of \(W\). Suppose that \(f(I) = 0\) and \(f(C) = 3\), where \(I\) is the \(2 \times 2\) identity matrix and \[ C = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}. \] Find \(f(B)\).
Solution
Write \[ A=\begin{bmatrix}a&b\\c&d\end{bmatrix}. \] Since \[ AB= \begin{bmatrix} 2a-b&-2a+b\\ 2c-d&-2c+d \end{bmatrix}, \] we have \(AB=0\) exactly when \(b=2a\) and \(d=2c\). Thus \[ W=\left\{\begin{bmatrix}a&2a\\c&2c\end{bmatrix}:a,c\in\RR\right\}. \] Let \[ f\!\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right) =pa+qb+rc+sd. \] Because \(f\) annihilates \(W\), \[ p+2q=0,\qquad r+2s=0. \] The conditions \(f(I)=0\) and \(f(C)=3\) give \[ p+s=0,\qquad s=3. \] Hence \[ (p,q,r,s)=\left(-3,\frac32,-6,3\right). \] Therefore \[ f(B)=2p-2q-r+s=-6-3+6+3=0. \]