Does a self adjoint matrix have an arbitrary $n^{th}$ root? Does a self adjoint matrix have an arbitrary $n^{th}$ root? i.e. a matrix $R$ such that $$R^n = N \space \forall \space n \in \mathbb{N}$$
I wrote out a proof showing that any self adjoint matrix $N$ has a square root that I am sure is very common using the spectral theorem, briefly, it runs like this:
Every self-adjoint matrix is normal, hence by the Spectral Theorem, there exists a unitary matrix U such that $ U^*NU = D $ is diagonal.
As $D$ has diagonal entries $d_1, ... d_n $, a new matrix, $S$, was defined with the square roots of the diagonal entries in $D$ on its diagonal, i.e. $s_i = \sqrt{d_i} $. Then $S^2 = D$ and defining $$ M = USU^* \rightarrow M^2 = US^2U^* = UDU^* $$
So $N$ has a square root $M$. I notice (I think) that this can be extended to cube roots, fourth roots and so on by replacing the square root with $s_i = d_i^{\frac{1}{n}} $ and going through basically the same argument. However, I think my argument is wrong as this does not seem very intuitive. What have I missed here?
 A: Your argument correctly shows that a normal matrix admits a $k$th root for any $k$. If there is anything here to give one pause, it's the fact $k$th root matrices, even for a self-adjoint matrix, can be highly non-unique, since any non-zero complex number (e.g., any non-zero eigenvalue of your matrix) admits $k$ $k$th roots. However, if your matrix is positive semidefinite, so that all the eigenvalues are non-negative, then you do have a unique positive semidefinite $k$th root for each $k$, which you get by taking the unique non-negative $k$th root of each eigenvalue, and plugging them into your construction.
Here's a higher-tech way of viewing things. Let $A$ be a normal matrix with unitary diagonalisation $$A = U \operatorname{diag}(\lambda_1,\dotsc,\lambda_n)U^\ast.$$ For any function $f : U \to \mathbb{C}$ defined on a set $U \subseteq \mathbb{C}$ containing all the eigenvalues of $A$, you can define the corresponding normal matrix $f(A)$ by $$f(A) = U \operatorname{diag}(f(\lambda_1),\dotsc,f(\lambda_n))U^\ast;$$ in particular, if $f(z)    =\sum_k a_k z^k$ is analytic on a disc about the origin of radius greater than $\|A\|$, the norm of $A$ with respect to your favourite matrix norm, then, indeed,
$$
 f(A) = \sum_k a_k A^k
$$
with absolute convergence in that matrix norm. The construction I just described is called the functional calculus in functional analysis. 
Now, suppose you want to find a $k$th root of $A$. This is the same as constructing $f(A)$ for $f(z) = z^{1/k}$, except that $f$ is multivalued, so you need to pick a branch of $f$. If, however, $A$ is positive semidefinite, then all the eigenvalues are non-negative, and hence any branch of $z^{1/k}$ restricting to the usual $k$-th root function $x \mapsto x^{1/k}$ on $[0,\infty)$ will give you the same result, namely, the unique positive semidefinite $k$th root of $A$. In short, the possible non-uniqueness of $k$th roots of $A$ is actually just a reflection of the non-uniqueness of branches of the multivalued function $f(z) = z^{1/k}$. 
A: This seems to be the way to go.
Actually, when talking about autoadjoint applications, the spectral theorem is used over and over again because it is so powerful.
