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Under what conditions does a matrix $A$ have a square root? I saw somewhere that this is true for Hermitian positive definite matrices(whose definition I just looked up).

Moreover, is it possible that for some subspace $X \subset M_n(\mathbb R)$ of $n\times n$ matrices over $\mathbb R$, the map $A \mapsto \sqrt{A}$ is continuous? People who want to consider more generality can also look at matrices over $\mathbb C$.

Thank you.

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    [A related question.](http://math.stackexchange.com/questions/57292) If your matrix possesses a Jordan block with a zero eigenvalue, you're shot. In general you need to peer at the Jordan form of your matrix and see if the square root function (or any other matrix function for that matter) is defined on your Jordan blocks.2011-09-17
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    @J.M.: Is looking at the Jordan form a continuous operation?2011-09-17
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    @J.M. That is not true. For all nilpotent Jordan blocks larger than $2\times 2$, squaring them yields a nontrivial endomorphism. This generates tons of nilpotent matrices which have square roots, even though no individual Jordan block has a square root. There is, in fact, a nice combinatorial description of the nilpotent matrices which have square roots, and which gives the Jordan form of all possible square roots (in general, there are several). Namely, if $a_i=\dim \ker A^i-\dim \ker A^{i-1}$, then $A$ will have a square root iff $a_i=a_{i+1}\Rightarrow a_i$ is even.2011-09-20
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    I suppose you're right, @Aaron, but would you happen to have an explicit example of a "rootable" nilpotent on hand, just so I can check my intuition? :)2011-09-20
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    @J.M. Consider the $4\times 4$ matrix with two $2\times 2$ blocks. This is the square of a matrix conjugate to a $4\times 4$ Jordan block, unless I've done my calculations wrong.2011-09-20

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