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In the case that the matrix A has non-zero eigenvalues $a_1$, ..., $a_n$ one may construct a square root $S$ as $$S=P(A)$$, where $P$ is a polynomial such that $$(d/dx)^n P(x) = (d/dx)^n \sqrt x$$ for all $x=a_k$ and $n\le d-1$, where $d$ is the dimension of the space. (Note: If one knows the maximal Jordan block sizes $b_1$,..., $b_n$ then one needs only to interpolate the square root up to derivatives of order $b_k-1$ at each eigenvalue. But you don't need to compute the jordan form or block sizes just to compute a square root.) This follows from the Jordan form, since the Jordan form tells us that for any polynomial $Q$ the value of $Q(A)$ depends only on the values of $Q$ and its derivatives up to $b_k-1$ at each of the eigenvalues of $A$. In particular, $$S^2=(P(A))^2=(P^2)(A)=A.$$ In the infinite-dimensional case one has the holomorphic calculus (also called the Dunford calculus), but in the finite-dimensional case one needs only polynomials.