Continuous square root function on the space of nxn matrices

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The discussion centers on the existence of a continuous square root function for n x n matrices, specifically whether such a function can be defined for all matrices or only within a limited region around the identity matrix. It is established that not all matrices have square roots, with an example provided of a non-square rootable matrix. Positive semidefinite matrices do have square roots, and the functional calculus is highlighted as a key method for constructing these roots. The conversation also touches on the existence of higher-order roots, questioning if similar limitations apply to cube roots and other powers. The need for further exploration of these concepts in functional analysis and spectral theory is emphasized.
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Homework Statement


Hi again all,
I've just managed to prove the existence (non-constructively) of a 'square root function' f on some open epsilon-ball about the identity matrix 'I' such that [f(A)]^2=A\qquad \forall\, A \,\text{ s.t.}\, \|I-A\|<\epsilon within Mn, the space of n*n matrices (note that's f(A)^2, not f^2(A), so for example the identity function wouldn't work) - I used the inverse function theorem on A^2 to deduce its existence. However, I was wondering whether there exists a function f such that f2(A)=A \forall A \in M_n? Or does there only exist such a function in a finite ball? What about for something like a cube root or a quintuple root function?

I was thinking perhaps some sort of compactness argument might work, but I couldn't reason anything in particular (and that's only if it isn't possible for the whole of M_n, otherwise compactness wouldn't work I don't suppose)

Many thanks for any help :)
 
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Any thoughts, anyone? Any help at all would be appreciated!
 
This is actually a fairly substantial question. Not every matrix has a square root; for instance, \begin{pmatrix}0&1\\0&0\end{pmatrix} does not. But positive semidefinite matrices do; there are several possible constructions. The most "fundamental" one from a functional analysis point of view uses something called the functional calculus, which makes it possible to "apply" many real functions, not just the square root function, to linear operators.

Look at the Wikipedia article "Square root of a matrix" to start. For a more pedagogical treatment, there is lots of good material about spectral theory in finite-dimensional spaces (which is what this is related to) in Paul Halmos's book Finite-dimensional vector spaces.
 
ystael said:
This is actually a fairly substantial question. Not every matrix has a square root; for instance, \begin{pmatrix}0&1\\0&0\end{pmatrix} does not. But positive semidefinite matrices do; there are several possible constructions. The most "fundamental" one from a functional analysis point of view uses something called the functional calculus, which makes it possible to "apply" many real functions, not just the square root function, to linear operators.

Look at the Wikipedia article "Square root of a matrix" to start. For a more pedagogical treatment, there is lots of good material about spectral theory in finite-dimensional spaces (which is what this is related to) in Paul Halmos's book Finite-dimensional vector spaces.

Thanks ever so much, how stupid of me not to spot that! Is it true of every order then, that there exists some matrix which isn't the n'th power of any matrix? (I couldn't come up with an example this late at night for the cube root, but I may be being slow ;-))
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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