Just a question out of curiosity: if you have a non-singular matrix, can you use the spectral decomposition and then take the square root of its eigen-values to get a square root of that matrix?
In order to show that each n*n nonsingular complex matrix has square root, I have shown that each of the Jordan blocks of its Jordan canonical form has square root. Now how can I show that the whole Jordan canonical form has square root?
If the matrix is not diagonalizable, then the knowing only the square roots of its eigenvalues does NOT tell you the square root of the matrix.
For example, the matrix has 1 as a double eigen value. In this particular case, it is easy to see that if is the square root of this matrix then
so that we must have , , , and .
Solving those, the two square roots are and . Knowing only that 1 is a double eigenvalue would not tell you that.