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?

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- Aug 7th 2013, 01:13 AMHumanSquare root of a 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?

- Aug 7th 2013, 05:13 AMchiroRe: Square root of a matrix
Hey Human.

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? - Aug 7th 2013, 06:49 AMHallsofIvyRe: Square root of a matrix
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.