Inverse of elementary matrix

Question

Let $Q^{p}_{q}\left( \lambda \right) = E_m + \lambda \left(0,...,e_p,...,0\right)$. How to show that $Q^{p}_{q}\left( \lambda \right)Q^{p}_{q}\left( - \lambda \right) = E_m$ holds true?

Answer 1

We assume that $p\neq q$. Otherwise, the statement no longer holds.

Note that we can write $Q^p_q(\lambda) = E_m + \lambda e_pe_q^T$. With that, we find $$ Q^p_q(\lambda) Q^p_q(-\lambda)= (E_m + \lambda e_pe_q^T)(E_m - \lambda e_pe_q^T) \\ =E_m + \lambda e_pe_q^T - \lambda e_p e_q^T + \lambda^2 (e_pe_q^T)(e_pe_q^T) \\= E_m + \lambda^2 (e_pe_q^T)(e_pe_q^T)\\ = E_m + \lambda^2 e_p(e_q^Te_p)e_q^T\\ = E_m + \lambda^2 e_p(0) e_q^T \\ = E_m $$ as was desired.