# 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? Commented Mar 30, 2014 at 13:59
• @GitGud Why wouldn't it be?
– fgp
Commented Mar 30, 2014 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. Commented Mar 30, 2014 at 14:03
• @user121692 No, it means it doesn't make sense. You need to define the symbol $A^{1/2}$. If you're just looking for a matrix $X$ such that $X^2=A$, then you should specify so, but this isn't always possible. Commented Mar 30, 2014 at 14:06
• @Git Gud Well for my specific purposes, the matrix I am working with are Markov chains transition matrix and the matrix A I wrote above is the annual transition probability. I need to calculate semi-annual, I suppose what I am looking for then is as you said $X$ such that $X^2 = A?$ Commented Mar 30, 2014 at 14:11

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? Commented Mar 30, 2014 at 14:01
• Sure, the matrix is diagonalizable, use $n=\frac{1}{2}$ and calculate. Commented Mar 30, 2014 at 14:02
• 'The'${{{{}}}}$? Commented Mar 30, 2014 at 14:10
• @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. Commented Mar 30, 2014 at 14:30
• Exactly. A calculator probably doesn't allow this operation on a matrix, but serious software for matrix manipulation swallows this without a problem. For instance, in octave/matlab A^(1/2) works directly (and gives the same result). And yes, just take the square root of the eigenvalues. Commented Mar 30, 2014 at 14:42