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Does a symmetric matrix necessarily have a symmetric square root, and why?

If not, then does a symmetric matrix that is also semi-definite necessarily have a symmetric square root (that may or may not be semi-definite), and why?

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2 Answers 2

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Any symmetric matrix $P$ (I assume you mean real matrices) can be diagonalized by an orthogonal matrix $U$: $$ P=U^T D U,\quad D=\left(\begin{array}{ccc} \lambda_1&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\lambda_n\\ \end{array}\right),\quad U U^T=I, $$ $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $P$, $\lambda_1,\ldots,\lambda_n\in\mathbb R$. If $P$ is positive semi-definite, then we have also $\lambda_1\ge 0,\ldots,\lambda_n\ge 0$. Consider the matrix $$ Q=U^T \left(\begin{array}{ccc} \sqrt{\lambda_1}&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\sqrt{\lambda_n}\\ \end{array}\right) U. $$ It is easy to see that $$ Q^2=U^T \left(\begin{array}{ccc} \sqrt{\lambda_1}&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\sqrt{\lambda_n}\\ \end{array}\right) U U^T \left(\begin{array}{ccc} \sqrt{\lambda_1}&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\sqrt{\lambda_n}\\ \end{array}\right) U= $$ $$ =U^T \left(\begin{array}{ccc} \sqrt{\lambda_1}&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\sqrt{\lambda_n}\\ \end{array}\right) I \left(\begin{array}{ccc} \sqrt{\lambda_1}&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\sqrt{\lambda_n}\\ \end{array}\right) U= U^T \left(\begin{array}{ccc} \lambda_1&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\lambda_n\\ \end{array}\right) U=P. $$

Now suppose that a symmetric matrix $R$ which is not positive semi-definite, i.e. has one or more negative eigenvalues, has a symmetric square root $S$. $S$ is diagonaliziable by an orthogonal matrix $U$, $$ S=U^T \left(\begin{array}{ccc} \lambda_1&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\lambda_n\\ \end{array}\right) U; $$ $$ S^2=U^T \left(\begin{array}{ccc} \lambda_1&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\lambda_n\\ \end{array}\right) U U^T \left(\begin{array}{ccc} \lambda_1&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\lambda_n\\ \end{array}\right) U= U^T \left(\begin{array}{ccc} \lambda_1^2&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\lambda_n^2\\ \end{array}\right) U. $$ We can see that the eigevalues of $R=S^2$ are $\lambda_1^2\ge 0,\ldots,\lambda_n^2\ge 0$. It contradicts with the assumption that $R$ has negative eigenvalues.

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  • $\begingroup$ Thanks! what exactly does the last argument prove? that a matrix that is symmetric but not positive semidefinite cannot have a symmetric square root? $\endgroup$
    – user56834
    Nov 13, 2017 at 7:48
  • $\begingroup$ @Programmer2134 Yes. In other words, the square of a symmetric matrix is always positive semi-definite $\endgroup$
    – AVK
    Nov 13, 2017 at 8:10
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Yes and no are the two answers. A symmetric matrix has a symmetric square root if and only if it is positive semi-definite. As simple proof is to use diagonalization.

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  • $\begingroup$ you mean no and yes? $\endgroup$
    – user56834
    Nov 13, 2017 at 6:19
  • $\begingroup$ No and yes is what I meant. Sorry! $\endgroup$ Nov 14, 2017 at 5:34

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