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

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  • $\begingroup$ Why do you even think it is defined? $\endgroup$
    – Git Gud
    Commented Mar 30, 2014 at 13:59
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    $\begingroup$ @GitGud Why wouldn't it be? $\endgroup$
    – fgp
    Commented Mar 30, 2014 at 14:01
  • $\begingroup$ @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. $\endgroup$
    – Git Gud
    Commented Mar 30, 2014 at 14:03
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    $\begingroup$ @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. $\endgroup$
    – Git Gud
    Commented Mar 30, 2014 at 14:06
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    $\begingroup$ @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 $X^2 = A?$ $\endgroup$
    – user121692
    Commented Mar 30, 2014 at 14:11

1 Answer 1

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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.

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  • $\begingroup$ Did you read the question? $\endgroup$
    – Git Gud
    Commented Mar 30, 2014 at 14:01
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    $\begingroup$ Sure, the matrix is diagonalizable, use $n=\frac{1}{2}$ and calculate. $\endgroup$
    – orion
    Commented Mar 30, 2014 at 14:02
  • $\begingroup$ 'The'${{{{}}}}$? $\endgroup$
    – Git Gud
    Commented Mar 30, 2014 at 14:10
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    $\begingroup$ @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. $\endgroup$
    – orion
    Commented Mar 30, 2014 at 14:30
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    $\begingroup$ 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. $\endgroup$
    – orion
    Commented Mar 30, 2014 at 14:42

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