Let \(M_{2\times2}\) denote the vector space of complex \(2\times2\) matrices. Let \[ A=\begin{pmatrix}0&1\\0&0\end{pmatrix} \] and let the linear transformation \(T:M_{2\times2}\to M_{2\times2}\) be defined by \(T(X)=XA-AX\). Find the Jordan canonical form for \(T\).
Solution
Let \(E_{ij}\) denote the standard matrix units in \(M_{2\times2}\).
1. The action of \(T\) on the standard basis is\[ \begin{aligned} T(E_{11})&=E_{12},& T(E_{12})&=0,\\ T(E_{21})&=E_{22}-E_{11},& T(E_{22})&=-E_{12}. \end{aligned} \]
Direct multiplication by \[ A=\begin{pmatrix}0&1\\0&0\end{pmatrix} \] gives \[ \begin{aligned} E_{11}A&=E_{12},& AE_{11}&=0,\\ E_{12}A&=0,& AE_{12}&=0,\\ E_{21}A&=E_{22},& AE_{21}&=E_{11},\\ E_{22}A&=0,& AE_{22}&=E_{12}. \end{aligned} \] Subtracting the right products from the left products gives the four displayed formulas.
2. Define\[ v_1=-2E_{12},\qquad v_2=E_{22}-E_{11},\qquad v_3=E_{21},\qquad v_4=E_{11}+E_{22}. \]Then\[ T(v_1)=0,\qquad T(v_2)=v_1,\qquad T(v_3)=v_2,\qquad T(v_4)=0. \]
By step 1, \[ \begin{aligned} T(v_1)&=-2T(E_{12})=0,\\ T(v_2)&=T(E_{22})-T(E_{11})=-2E_{12}=v_1,\\ T(v_3)&=T(E_{21})=E_{22}-E_{11}=v_2,\\ T(v_4)&=T(E_{11})+T(E_{22})=0. \end{aligned} \]
3. The ordered family \((v_1,v_2,v_3,v_4)\) is a basis of \(M_{2\times2}\).
Suppose \[ a v_1+b v_2+c v_3+d v_4=0. \] In the basis \((E_{11},E_{12},E_{21},E_{22})\), the left-hand side is \[ (-b+d)E_{11}-2aE_{12}+cE_{21}+(b+d)E_{22}. \] Hence \[ -b+d=0,\qquad -2a=0,\qquad c=0,\qquad b+d=0. \] Thus \(a=b=c=d=0\), so the four vectors are linearly independent. Since \(M_{2\times2}\) has dimension \(4\), they form a basis.
4. The Jordan canonical form of \(T\) is\[ \boxed{ J_3(0)\oplus J_1(0) = \begin{pmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix} }. \]
By step 3, \((v_1,v_2,v_3,v_4)\) is a basis. Step 2 shows that, in this basis, \(v_3\mapsto v_2\mapsto v_1\mapsto0\) is one Jordan chain of length \(3\), while \(v_4\mapsto0\) is a Jordan chain of length \(1\). Therefore the matrix of \(T\) in this basis is exactly the displayed block diagonal Jordan matrix.
5. Q.E.D.
Step 4 gives the required Jordan canonical form.