Let \(A,B\) be real \(2\times2\) matrices satisfying \[ A^2=B^2=I, \qquad AB+BA=0. \] Show that there exists a real invertible \(2\times2\) matrix \(T\) such that \[ TAT^{-1}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad TBT^{-1}=\begin{pmatrix}0&1\\1&0\end{pmatrix}. \]
Solution
1. The matrix \(A\) is diagonalizable over \(\RR\), with eigenvalues among\(\{1,-1\}\).
The relation \[ A^2=I \] shows that the minimal polynomial of \(A\) divides \[ t^2-1=(t-1)(t+1). \] This polynomial splits over \(\RR\) with distinct roots, so \(A\) is diagonalizable and its eigenvalues lie in \(\{1,-1\}\).
2. Both \(1\) and \(-1\) occur as eigenvalues of \(A\).
If \(A=I\), then \[ AB+BA=2B=0, \] so \(B=0\), contradicting \(B^2=I\). The same contradiction follows if \(A=-I\). Since step 1 makes \(A\) diagonalizable with eigenvalues only \(\pm1\), it follows that both eigenvalues occur.
3. Choose a nonzero vector \(v\) with\[ Av=v. \]Then \(Bv\) is a nonzero eigenvector of \(A\) with eigenvalue \(-1\).
The relation \[ AB=-BA \] gives \[ A(Bv) = -B(Av) = -Bv. \] Moreover, \(Bv\neq0\) because \(B^2=I\) makes \(B\) invertible.
4. The vectors\[ v,\ Bv \]form a basis of \(\RR^2\).
By step 3, they are nonzero eigenvectors of \(A\) for the distinct eigenvalues \(1\) and \(-1\). Hence they are linearly independent. Since the space has dimension \(2\), they form a basis.
5. In the ordered basis \((v,Bv)\), the matrices of \(A\) and \(B\) are\[ \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \qquad\text{and}\qquad \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}, \]respectively.
For \(A\), step 3 gives \[ Av=v, \qquad A(Bv)=-Bv. \] For \(B\), \[ Bv=Bv \] is the second basis vector, while \[ B(Bv) = B^2v = v \] is the first basis vector. These coordinate actions give the two displayed matrices.
6. There is an invertible real matrix \(T\) satisfying the two requiredconjugation identities.
Let \[ P= \begin{pmatrix} \vert&\vert\\ v&Bv\\ \vert&\vert \end{pmatrix}. \] Step 4 makes \(P\) invertible. Step 5 says \[ P^{-1}AP = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \] and \[ P^{-1}BP = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}. \] Taking \[ T=P^{-1} \] gives exactly the required formulas.
7. Q.E.D.
Step 6 constructs the required real invertible matrix \(T\).