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Does a real, non-symmetric, non-negative definite matrix have a unique non-negative definite square root? If yes, is taking this square root a continuous map?

The reason I'm thinking intuitively yes is that, we know that symmetric non-negative definite matrices have unique non-negative definite square roots. However, in the proofs, we heavily rely on the diagonalization, but not orthogonal diagonalization for a real symmetric matric. That is, if we've a matrix $M$ so that $M = P^{-1}DP$ for dome diagonal matrix $D$, then we have the eigenvalues of $D$ non negative. So it makes sense to define $\sqrt D $, and hence $\sqrt(M):=P^{-1}\sqrt D P$. Note don't need $P$ to be orthogonal to define $\sqrt M$.

So now, if we don't assume orthogonality, can every matrix $M$, that're non-symmetric, non-diagonalizable, have a unique non-symmetric square root $N$ so that $N^2=M?$ Note that, of course we don't want $M=NN^{T}$, as then $M$ will be forced to be symemtric.

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    $\begingroup$ Jordan normal form (en.wikipedia.org/wiki/Jordan_normal_form) may be of help, the square root is then the taylor expension of the function $\sqrt{}$ applied to the jordan matrix with same unitary matrices on the left and on the right $\endgroup$
    – P. Quinton
    Nov 19, 2019 at 12:55
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    $\begingroup$ The answer is yes. I have yet to see a proof for the positive definite case that fails to apply to the non-negative definite case, so I'm not sure how you can deduce one without deducing the other. $\endgroup$ Nov 19, 2019 at 13:08
  • $\begingroup$ @Omnomnomnom Thanks for your comment, that's what I also thought, but I can clearly see that non negative real symmetric matrices have unique non negative square root; here the fact that real symmetric matrices are diagonalizable (in fact, orthogonally diagonalizable) comes into play. But what if the non-negative matrix is not diagonalizable? For example, math.stackexchange.com/questions/473118/… has an answer that gives an example of a non-symmetric positive, non-diagonalizable matrix. (contd) $\endgroup$ Nov 19, 2019 at 14:25
  • $\begingroup$ @Omnomnomnom (contd.) I'm trying to drop the symmetry condition, and see if the non-symetric positive definite/non-ive definite matruces have positive/non-ive definite square roots. Again, if we assume they're (not nrcessary orthogonally) diagonalizable, then the previous argument in the previous paragraph applies. But what if not diagonilable at all? Fr example, does the positive definite matrix in math.stackexchange.com/questions/473118/… have a square root? $\endgroup$ Nov 19, 2019 at 14:28
  • $\begingroup$ @P.Quinton Thank you for your comment, I'm not clear yet. do you mind writing a detailed answer please? P.S. the matrices I'm considering aren't diagonizable, for if they're, then we definitely have their unique non-ive definite square roots. $\endgroup$ Nov 19, 2019 at 15:14

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For any complex valued square matrix $M$, you can find the Jordan normal form $M=P^{-1} J P$. Suppose that $0$ is an eigenvalue of $M$ of multiplicity $2$ or more. Then applying $\sqrt{}$ to $M$ is not possible since we need to evaluate the derivative of $\sqrt{}$ at $0$ for the block corresponding to the $0$ eigen value.

This is not an argument against existence of the square root but it may be an intuition on what may go wrong. If the multiplicity of $0$ is strictly less than $2$, then the square exists by the square root of $M$.

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  • $\begingroup$ A simple example is $\begin{bmatrix} 0 & 0 \\ 0& 0 \end{bmatrix}$ and $\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}$ and $\begin{bmatrix} 0 & 0 \\ 1& 0 \end{bmatrix}$ are non-negative definite square roots of 0. $\endgroup$
    – N. S.
    Nov 19, 2019 at 16:22
  • $\begingroup$ I feel like there is relation between the multiplicity $0$ and the dimension of the space of solution of $M=N^2$ $\endgroup$
    – P. Quinton
    Nov 19, 2019 at 16:26
  • $\begingroup$ @N.S. I interpret "non-negative definite" as "positive semidefinite", and the only positive semidefinite square root of $0 \in M_2(\mathbb{R})$ is $0$. The quadratic form associated to a square root of $0$ is [equivalent to] $(x,y) \mapsto c\cdot xy$ for some $c$, and unless $c = 0$ that's indefinite. $\endgroup$ Nov 19, 2019 at 16:51

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