Let \(A,B\) be real \(n\times n\) matrices such that \[ A^3=B^5=I_n,\qquad AB=BA. \] Show that \(A+B\) is invertible.
Solution
1. Let \(v\in\RR^n\) satisfy\[ (A+B)v=0. \]Then\[ A^2v=B^2v \]and\[ A^3v=-B^3v. \]
The kernel assumption gives \[ Av=-Bv. \] Using \(AB=BA\), \[ A^2v = A(-Bv) = -ABv = -BAv = B^2v. \] Applying \(A\) once more and commuting it past \(B^2\) gives \[ A^3v = A(B^2v) = B^2Av = -B^3v. \]
2. The vector \(v\) satisfies\[ B^3v=-v \]and\[ B^2v=-v. \]
Since \(A^3=I_n\), step 1 gives \[ v=A^3v=-B^3v, \] so \(B^3v=-v\). Applying \(B^2\) to this equality and using \(B^5=I_n\) gives \[ v = B^5v = -B^2v, \] hence \(B^2v=-v\).
3. One has \(v=0\).
Apply \(B\) to the equality \(B^2v=-v\) from step 2: \[ B^3v=-Bv. \] The other equality in step 2 gives \(B^3v=-v\), so \[ Bv=v. \] Applying \(B\) once more yields \[ B^2v=v. \] But step 2 also gives \(B^2v=-v\). Therefore \[ v=-v. \] Over \(\RR\) this forces \(v=0\).
4. The matrix \(A+B\) is invertible.
Step 3 shows \[ \ker(A+B)=\{0\}. \] Thus the linear endomorphism of the finite-dimensional space \(\RR^n\) defined by \(A+B\) is injective, hence bijective. Therefore \(A+B\) is invertible.
5. Q.E.D.
Step 4 is the required conclusion.