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One does not actually need the Jordan form to construct a square root in the case of non-zero eigenvalues, although simply knowing that it exists is quite helpful. Let $a_1$, ..., $a_n$ denote the non-zero complex eigenvalues. Then a square root $S$ is given by $$S=P(A)$$, where $P$ is any 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. 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 order $b_k-1\le n-1$ at each of the eigenvalues of $A$, where $b_k$ is the size of the maximal Jordan block for the corresponding eigenvalue $a_k$. 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.