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.
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 $M$.