Let \(C^{2 \times 2}\) be the complex vector space of \(2 \times 2\) matrices with complex entries. Let \[ B = \begin{bmatrix} 1 & -1 \\ -4 & 4 \end{bmatrix} \] and let \(T\) be the linear operator on \(C^{2 \times 2}\) defined by \(T(A) = BA\). What is the rank of \(T\)? Can you describe \(T^2\)?
Solution
The matrix \[ B=\begin{bmatrix}1&-1\\-4&4\end{bmatrix} \] has rank \(1\): its second row is \(-4\) times its first row, and \(B\ne0\). For \(A=[a_1\ a_2]\), with columns \(a_1,a_2\in\mathbb C^2\), \[ T(A)=BA=[Ba_1\ Ba_2]. \] Each column \(Ba_j\) can be any vector in the one-dimensional space \(\operatorname{im}B\), independently of the other column. Hence \[ \operatorname{range}T =\{[u\ v]:u,v\in\operatorname{im}B\} \] has dimension \(2\). Therefore \(\operatorname{rank}T=2\).
A direct multiplication gives \[ B^2= \begin{bmatrix}5&-5\\-20&20\end{bmatrix}=5B. \] Consequently, for every \(A\), \[ T^2(A)=B(BA)=B^2A=5BA=5T(A). \] Thus \(T^2=5T\).