Suppose that \(A\) is an \(m\) by \(n\) complex matrix and \(B\) is an \(n\) by \(m\) complex matrix, and write \(I_m\) for the \(m\) by \(m\) identity matrix. Show that if \(I_m-AB\) is invertible then so is \(I_n-BA\). (Hint: what does the condition that \(I_m-X\) is not invertible say about eigenvalues and eigenvectors of \(X\)?)
Solution
Assume that \(I_m-AB\) is invertible.
1. The kernel of \(I_n-BA\) is trivial.
Let \(v\in\CC^n\) satisfy \[ (I_n-BA)v=0. \] Then \(BAv=v\). Set \(w\coloneqq Av\in\CC^m\). We have \[ ABw = ABAv = A(BAv) = Av = w, \] so \((I_m-AB)w=0\). Since \(I_m-AB\) is invertible, \(w=0\). Therefore \[ v=BAv=Bw=0. \] Thus \(\ker(I_n-BA)=\{0\}\).
2. The matrix \(I_n-BA\) is invertible.
By step 1, the linear map \(I_n-BA:\CC^n\to\CC^n\) is injective. An injective endomorphism of the finite-dimensional vector space \(\CC^n\) is surjective, hence invertible.
3. Q.E.D.
Step 2 is the required conclusion.