Let \(A\) and \(B\) be real \(2\times2\) matrices such that \(A^2=B^2=I\) and \(AB+BA=0\). Prove there exists a real nonsingular matrix \(T\) with
\[ TAT^{-1}=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad TBT^{-1}=\begin{pmatrix}0&1\\1&0\end{pmatrix}. \]
Solution
1. The matrix \(A\) is diagonalizable over \(\RR\), and both \(1\) and \(-1\) areeigenvalues of \(A\), each with a one-dimensional eigenspace.
Because \(A^2=I\), the minimal polynomial of \(A\) divides \((t-1)(t+1)\), which has distinct real roots. Hence \(A\) is diagonalizable over \(\RR\) with eigenvalues in \(\{1,-1\}\).
If \(A=I\), then \[ AB+BA=2B=0, \] which is impossible because \(B^2=I\) makes \(B\) invertible. Similarly, \(A=-I\) would give \[ AB+BA=-2B=0, \] again impossible. A diagonalizable \(2\times2\) matrix with eigenvalues in \(\{1,-1\}\) that is neither \(I\) nor \(-I\) has both eigenvalues, each with a one-dimensional eigenspace.
2. \(B\) maps the \(\lambda\)-eigenspace of \(A\) isomorphically onto the\((-\lambda)\)-eigenspace, for \(\lambda\in\{1,-1\}\).
Let \(v\) satisfy \[ Av=\lambda v. \] From \(AB=-BA\), \[ A(Bv)=-B(Av)=-\lambda Bv. \] Because \(B\) is invertible, \(Bv\ne0\) whenever \(v\ne0\), and \(B\) is injective between these one-dimensional eigenspaces.
3. Let \(u\ne0\) satisfy \(Au=u\), and let \(P\) be the real matrix withcolumns \(u\) and \(Bu\). Then \(P\) is nonsingular and\[ P^{-1}AP=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad P^{-1}BP=\begin{pmatrix}0&1\\1&0\end{pmatrix}. \]
By step 2, \[ A(Bu)=-Bu, \] so \(u\) and \(Bu\) are nonzero vectors in distinct eigenspaces of \(A\) and form a basis of \(\RR^2\). Hence \(P\) is nonsingular. In this basis, \[ Au=u, \qquad A(Bu)=-Bu, \] which gives the matrix of \(A\), and \[ B(u)=Bu, \qquad B(Bu)=B^2u=u, \] which gives the matrix of \(B\).
4. The real nonsingular matrix\[ \boxed{T=P^{-1}} \]satisfies\[ TAT^{-1}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad TBT^{-1}=\begin{pmatrix}0&1\\1&0\end{pmatrix}. \]
With \(T=P^{-1}\), one has \(TAT^{-1}=P^{-1}AP\) and \(TBT^{-1}=P^{-1}BP\); step 3 computes both.
5. Q.E.D.
Step 4 exhibits the required matrix \(T\).