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?
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 $\displaystyle \begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$ has 1 as a double eigen value. In this particular case, it is easy to see that if $\displaystyle \begin{bmatrix}a & b \\ c & d \end{bmatrix}$ is the square root of this matrix then
$\displaystyle \begin{bmatrix}a^2+ bc & ab+ bd \\ ac+ cd & bc+ d^2\end{bmatrix}= \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$
so that we must have $\displaystyle a^2 +bc= 1$, $\displaystyle ab+ bd= 1$, $\displaystyle ac+ cd= 0$, and $\displaystyle bc+ d^2= 1$.
Solving those, the two square roots are $\displaystyle \begin{bmatrix}1 & 1/2 \\ 0 & 1\end{bmatrix}$ and $\displaystyle \begin{bmatrix}-1 & -1/2 \\ 0 & -1\end{bmatrix}$. Knowing only that 1 is a double eigenvalue would not tell you that.