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I have a square matrix called A. How can I find $A ^ {-1/2}$. Should I compute $a_{ij} ^ {-1/2}$ for all of its elements?

Thanks

Taban
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    What properties has $A$? Not all matrices have a square root and when it has one there is often non-uniqueness, one has to ask for properties satisfied by the square root. Typically, if $A$ is symmetric definite positive, $A$ has a unique positive definite square root. – C. Falcon Jun 27 '16 at 18:26
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    Try that construction for a 2-by-2 example and see if $(A^{-1/2})^2=A^{-1}$. If it doesn't, then that's not $A^{-1/2}$. – Semiclassical Jun 27 '16 at 18:27

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$A=SDS^{−1}$, and $A^k=SD^kS^{−1}$ where D is the diagonal matrix composed from eigenvalues, S is the matrix of eigenvectors, and $S^{−1}$ is the inverse of S.
So the first step is to find the eigenvalues, and then find the corresponding eigenvectors.
Then use $A^{1/2}=SD^{1/2}S^{-1}$, and finally find the inverse $A^{-1/2}=(A^{1/2})^{-1}$

Valeria
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Firstly you will need to pick a branch of the square root function over the field of which the elements of your matrix belong. Once you have done that,

  1. If $A$ is diagonalizable you can do as @Annalise writes.

  2. If $A$ is orthogonally diagonalizable you can do as @Junning Li writes.

However for example if $A$ is not diagonalizable, then it can maybe still be put on some canonical form:

$$A = TCT^{-1}$$

Where $C$ can be block-diagonal matrix. In this case we can do approximate square root by trying some power series expansion on the diagonal blocks of $C$. However this is not guaranteed to make any sense. It will depend a lot on application.

mathreadler
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It exists only if $A$ is invertible, i.e. $A^{-1}$ exists. Now this matrix have to follow certain properties , then a square root can be computed.

You always have the EVD at hand. Break the matrix $A^{-1}=U\cdot \Sigma\cdot U^t$ Where $U$ is a unitary matrix.Check that $$(U\cdot \Sigma^{1/2}\cdot U^t)^2=U\cdot \Sigma^{1/2}\cdot U^t\cdot U\cdot \Sigma^{1/2}\cdot U^t=A^{-1}$$ And $[\Sigma^k]_{ii}=[\Sigma_{ii}^k]$

Community
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Qwerty
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Eigen-decomposition: $A = U * \Lambda * U^t$.

$A^{-1/2} = U * \Lambda^{-1/2} * U^t$.

This method is not a very good for numerical computation.

Junning Li
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