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this comes from a comment of this where the comment use $A^{1/2}$ ,

I learned that a integer kth power of a matrix, if A can be diagonizled as $XDX^{-1}$ then $A^k=XD^kX^{-1}$, even when k<=0

so the OP means $A^{1/2}=XD^{1/2}X^{-1}$?

what's a general definition of matrix's exponent if the exponent is real or complex?

femto
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  • This is well defined for symmetric definite positive matrices, you have an uniqueness of the square root, otherwise I suggest you use power expansion with $(I + (A-I))$ but you have to verify $|A-I|_2 < 1$, you definition would do the job though – julio_es_sui_glace Mar 19 '24 at 10:47

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Th problem with square root for matrices is the lack of uniqueness, you don't have an integral ring, hence you can have infinite roots for the polynomial $X^2 -A$. For example $X^2 = 0$ in $\mathcal{M}_2(\mathbb{R})$ has all nilpotent matrices as solution (and $0$ is diagonalisable).

One way to achieve uniqueness is to work with symmetric definite positive matrices, which have an unique square root.

Otherwise you could use for $||| A-I||| < 1$: $$A^{1/2} = (I + (A-I))^{1/2} = \sum_{k\geq 0} \binom{1/2}{k}A^k.$$ This gives you a (not unique) but canonical square root.

You solution would work and would be a canonical way to find a square root, to detail it, it is better to work with the eigenspaces, if you have $\lambda_1,\dots,\lambda_d$ as eigenvalues, you define $A_{|E_\lambda} = \lambda^{1/2} id$, but think about it, the uniqueness of the square root is not even verified in $\mathbb{C}$ and if we were to choose a canonical root, we would have to do it on $\mathbb{C} \setminus \mathbb{R}_-$, so you have to ensure something on the eigenvalues.

One last solution is to work with the D+N decomposition, (or dunford decomposition) in case you don't have a diagonalisable matrix but you won't find a canonical way to find it.

If you think about it, you would very much like to find a square root that is a polynomial in $A$ so finding a root in $k[A]$ would be equivalent to finding a root of $\bar{X}$ in $k[X]/(\pi_A)$

julio_es_sui_glace
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