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If I want to take the power of matrix $A$ with e.g 3, $A^3$ or with power of $-\frac {1}{2}$, e.g $A^{-\frac {1}{2}}$ etc.

Is there an easy way to solve $A^n$, where $n\in R$ and $A \in R^{nxn}$ by using QR, LU, SVD, EIG or other decompositions?

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  • $\begingroup$ Robert's answer also works with Jordan form, $A = S J S^{-1}.$ the detail is that the diagonal part of $J$ commutes with the nilpotent part (the off-diagonal $1$s) $\endgroup$ – Will Jagy Feb 10 at 2:19
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If you can diagonalize the matrix as $A = S D S^{-1}$, then $A^n = S D^n S^{-1}$, where for $D^n$ you just need to take the $n$'th power of the diagonal elements of $D$.

Caution: if $n$ is not an integer, the $n$'th power is multivalued over the complex numbers. This will even affect real matrices which have non-real eigenvalues. Also, $n$ is not an integer and the eigenvalues are not all distinct, there may be other $n$'th powers not of this form.

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  • $\begingroup$ How do I find S and D ? Yes. N can be a decimal value. $\endgroup$ – Daniel Mårtensson Feb 10 at 11:11
  • $\begingroup$ Most important are integer values, and square root and also negative square root. $\endgroup$ – Daniel Mårtensson Feb 10 at 11:20
  • $\begingroup$ Diagonalization $\endgroup$ – Robert Israel Feb 10 at 14:52
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In case $A$ is diagonizable with a Diagonal matrix $D$ (having the eigenvalues on its diagonal and a matrix H, such that $A=H^{-1}DH$, then $A^n = H^{-1}D^nH$.

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