# Fractional power of matrix

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 $A^{\frac{1}{2}}, A^{\frac{3}{2}}$?

• Why do you even think it is defined? – Git Gud Mar 30 '14 at 13:59
• @GitGud Why wouldn't it be? – fgp Mar 30 '14 at 14:01
• @fgp Why would it be? Matrices necessarily having square roots? Square roots over what field? Furthermore the square root need not be unique, so using the symbol $\sqrt A$ leads to problems. – Git Gud Mar 30 '14 at 14:03
• And for what's it's worth, this is a stochastic matrix, so non-real solutions are probably not acceptable. – Git Gud Mar 30 '14 at 14:04
• I tried typing this into my graphing calculator and I am getting an error, so does this mean the answer is not unique? – user121692 Mar 30 '14 at 14:05

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.

• Did you read the question? – Git Gud Mar 30 '14 at 14:01
• Sure, the matrix is diagonalizable, use $n=\frac{1}{2}$ and calculate. – orion Mar 30 '14 at 14:02
• 'The'${{{{}}}}$? – Git Gud Mar 30 '14 at 14:10
• If there are 8 "square roots", is only one of them a valid transition matrix? – Neil W Mar 30 '14 at 14:25
• @Neil the same applies as for reals: you define it to be the one with positive eigenvalues. That's why you start having problems selecting the branch when the original matrix is not positive definite. – orion Mar 30 '14 at 14:30