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I know the answer is No, since you can put plus/minus on each eigenvalue. But how about putting a psd requirement? Like $A = S^2$, $S$ is psd, is $S$ unique?

I was worried about the case where if $\lambda_1 = \lambda_2$, then we can flip the first two columns of unitary matrix $P$. You can require all eigen values distinct. But even if has same eigen values, the matrix shall be the same?

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  • $\begingroup$ Are you assuming that $A$ is PSD? Otherwise you can't find a PSD square root. If so, then yes, there is a unique PSD square root. $\endgroup$
    – Federico Poloni
    Commented Apr 3, 2017 at 19:23
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    $\begingroup$ PSD = positive semi-definite, for those like me who need to google to decipher the abbreviations (possibly standard in English-speaking education, not mine) $\endgroup$
    – YCor
    Commented Apr 3, 2017 at 19:30
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    $\begingroup$ Answer first the easier problem: how may psd square roots does the identity matrix have? $\endgroup$
    – Liviu Nicolaescu
    Commented Apr 3, 2017 at 19:40

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A positive semidefinite matrix has a unique PSD square root - Horn&Johnson Theorem 7.2.6. Much more is know, see, for example,

Johnson, Charles R.; Okubo, Kazuyoshi; Reams, Robert, Uniqueness of matrix square roots and an application, Linear Algebra Appl. 323, No.1-3, 51-60 (2001). ZBL0976.15009. (available for free on line)

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