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I know that if $A$ is a square hermitian matrix positive definite (positive semi-definite) then $A^{1/2}$ exists and is also a square hermitian matrix positive definite (positive semi-definite respectively). However I want to know how weaker can I make the hypothesis for the existance of $A^{1/2}$. Does it work for any square matrix? Do I need $A$ to be at least positive semi-definite?

leplata
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    @user not true. There are no square-root of $\begin{pmatrix}0&1\0&0\end{pmatrix}$. – user10354138 Feb 24 '21 at 02:39
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    Have a look at https://math.stackexchange.com/questions/65227/under-what-conditions-does-a-matrix-a-have-a-square-root?rq=1 – 1Rock Feb 24 '21 at 02:52

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If $A$ is diagonalizable, $QAQ^{-1} = D$ with $D = \mathrm{diag}(d_1,\ldots,d_n)$; and each $d_i$ has a square root in the field you are considering (for example, this is always the case in $\mathbb C$), then yes: take $D' = \mathrm{diag}(s_1,\ldots,s_n)$ with $s_i^2 = d_i$ and then

$$ (Q^{-1}D'Q)^2 = Q^{-1}D'^2Q = Q^{-1}DQ = A. $$

Note that, however, a square root in general is not unique. Moreover, without any hypothesis they may fail to exist.

qualcuno
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