Use the result of Exercise 15 to prove that if \(c\) is a non-zero complex number and \(N\) is a nilpotent complex matrix, then \((cI + N)\) has a square root. Now use the Jordan form to prove that every non-singular complex \(n \times n\) matrix has a square root.
Solution
1. Every block \(cI+N\) with \(c\ne0\) and \(N\) nilpotent has a square root.
Choose \(s\in\mathbb C\) with \(s^2=c\). Since \(c\ne0\), the matrix \[ M=c^{-1}N \] is nilpotent. By Exercise 15 there is a polynomial \(R=q(M)\) such that \[ R^2=I+M \] [@HK71]. Put \(B=sR\). Then \[ B^2=s^2R^2 =c(I+c^{-1}N) =cI+N. \] Thus every such shifted nilpotent block has a square root.
2. Every nonsingular complex matrix is assembled from such blocks.
Let \(A\) be nonsingular. Choose a Jordan form \[ A=PJP^{-1}, \qquad J=\bigoplus_{\ell}J_{k_\ell}(c_\ell). \] Nonsingularity means \(0\) is not an eigenvalue, so each \(c_\ell\ne0\). Every Jordan block has the form \[ J_{k_\ell}(c_\ell)=c_\ell I+N_\ell \] with \(N_\ell\) nilpotent. By step 1 choose \(B_\ell\) satisfying \(B_\ell^2=J_{k_\ell}(c_\ell)\). Set \[ B=\bigoplus_\ell B_\ell. \] Then \(B^2=J\). Therefore \[ X=PBP^{-1} \] satisfies \[ X^2=PB^2P^{-1}=PJP^{-1}=A. \] Hence every nonsingular complex matrix has a square root.