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

Git Gud
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user121692
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  • Why do you even think it is defined? – Git Gud Mar 30 '14 at 13:59
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    @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
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    @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. – Git Gud Mar 30 '14 at 14:06
  • @GitGud Well, you don't need a field to define square roots - for every ring $R$ there's the ring of polynomials $R[x]$ and you can thus ask for zeros of $s_a(x) = x^2 - a$ for any $a \in R$. Thus, $A^\frac{p}{q}$ is naturally interpreted as "find a matrix $B$ such that $A^p = B^q$". Whether or not there's a unique solution is a different question... – fgp Mar 30 '14 at 14:08
  • @fgp The field thing was just to set a background. And no, you cannot define a symbol as having more than one meaning in a fixed context. – Git Gud Mar 30 '14 at 14:10
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    @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 $X^2 = A?$ – user121692 Mar 30 '14 at 14:11
  • @user121692 OK, that has been answered. It's possible here. – Git Gud Mar 30 '14 at 14:11

1 Answers1

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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.

orion
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