TITLE: Fractional power of matrix
QUESTION [12 upvotes]: If I am given a matrix, for example $A = \begin{bmatrix}
       0.7 & 0.2 & 0.1           \\[0.3em]
       0.2 & 0.5           & 0.3 \\[0.3em]
       0           & 0 & 1
     \end{bmatrix}$,
how do I calculate the fractional matrix like $A^{\frac{1}{2}}, A^{\frac{3}{2}}$?

REPLY [16 votes]: If a matrix is diagonalizable, then diagonalize it, $A=PDP^{-1}$ and apply the power to the diagonal
$$A^n=PD^n P^{-1}$$
The diagonal values are acted on individually.

octave gives:
$$P=\begin{bmatrix}   0.85065 & -0.52573 &  0.57735\\
   0.52573 &  0.85065 &  0.57735\\
   0.00000 &  0.00000 &  0.57735\end{bmatrix}$$
$$D=\operatorname{diag}(0.82361, 0.37639,1)$$
I realize this is a numerical uglyness but I don't have a symbolic manipulation software at hand from this computer. However, the eigenvalues are different so this is a diagonalization.
The square root is
$$\sqrt{A}= \begin{bmatrix}0.82626 & 0.13149 & 0.04225\\
   0.13149 &  0.69477 &  0.17374\\
   0.00000 &  0.00000  & 1.00000\end{bmatrix}$$

This definition satisfies the requirement for roots that $(A^{1/p})^p=A$ for positive definite matrices (just like with $\sqrt{x}$ for scalars).
In a similar way, you can define functions on matrices through their power series. For instance, $e^A=P \exp(D)P^{-1}$ is perfectly well defined for diagonalizable matrices.
The convergence criteria and domain of these functions gets generalized and usually involves conditions for eigenvalues, positive-definiteness, symmetry, ortogonality and so on.
Note that the term square root of a matrix is sometimes used to represent a Cholesky decomposition, which instead works as $A=LL^T$ where $L$ is a lower triangular matrix. This is not the square root in the strictest sense, but it works like one for some numerical procedures.