The following problems provide a way to find the squareroot

The following problems provide a way to find the squareroot of a general 3 times 3 non-singular matrix over C (we already know how to do this for self-adjoint matrices using diagonalization). Definition. A square matrix N is called nilpotent. if there exists some integer k > 0, such that N^k = 0. If N is a n times n nilpotent matrix, then its minimal polynomial must divide x^k, and so its characteristic polynomial must be x^n. Thus by Cayley-Hamilton theorem. N^n = 0. This fact may be useful in the following problems. If N is a nilpotent 3 times 3 matrix over C. prove that the matrix A = I + 1/2N-1/8N^2    (1) satisfies A^2 = I + N, i.e., A is a square unit of I + N. Then prove that for any lambda Notequalto 0, lambdaI + N has a square wot as well. Use the .Ionian normal form to prove that every complex 3 times 3 non-singular matrix A (i.e. det A Notequalto 0) has a squareroot Remark. This result actually holds for any complex n times n non-singular matrix A. The key step is that for any n times n nilpotent matrix N. bv the Taylor series of (1 + t)^1/2 = 1 + 1/2 t - 1/8 t^2 + 1/16 t^3 + ...., we can obtain a formula for a squareroot of I + N, similar to (1), but with n + 1 terms.

Solution

A = I + N/2 -N^2/8

A^2 = I + N^2/4 + N^4/64 + N - N^3/8 - N^2/4

Since N^3 = 0

this gives

A^2 = I + N