The principal matrix logarithm and square root exist off the closed negative-real axis

On a Jordan block $J=\lambda I+N$ of size $d$, the primary value of a holomorphic $h$ is ${\sum }_{k<d}{h}^{(k)}(\lambda )N^k\mathrm{/}k!$. Products of such truncated Taylor polynomials agree modulo $N^d$ with the truncated Taylor polynomial of the scalar product, by the finite Leibniz rule. For composition $h\circ g$, assume $g$ is holomorphic near $\lambda$ and $h$ near $g(\lambda )$. Write $g(J)=g(\lambda )I+V$, where $V={\sum }_{j=1}^{d-1}{g}^{(j)}(\lambda )N^j\mathrm{/}j!$ and $V^d=0$. Every Jordan block of $g(J)$ therefore has eigenvalue $g(\lambda )$ and size at most $d$, so [F3] gives $h(g(J))={\sum }_{k=0}^{d-1}{h}^{(k)}(g(\lambda ))V^k\mathrm{/}k!$. Substituting the finite jet for $V$ yields precisely the Taylor jet of $h\circ g$ at $\lambda$ modulo $N^d$, since every discarded term has degree at least $d$. Thus analytic identities on a neighborhood of $\lambda$ remain valid on this Jordan block. In particular, the matrix exponential series agrees with the primary value of the scalar exponential: the binomial theorem and absolute convergence in [F5] give ${\sum }_{n\ge 0}\frac{J^n}{n!}={\sum }_{k=0}^{d-1}\frac{N^k}{k!}\left({\sum }_{r\ge 0}\frac{{\lambda }^r}{r!}\right)={\sum }_{k=0}^{d-1}\frac{e^{\lambda }{N}^k}{k!}\mathrm{.}$ Only finitely many nilpotent powers occur, so rearranging the absolutely convergent sum is legitimate.

Applying step 1.1 to $\exp \circ \mathrm{Log}(z)=z$ and $(z_{\mathrm{Log}}^{}{)}^2=z$ gives $\exp (L(A))=A$ and $R(A{)}^2=A$ block by block and then by similarity. The eigenvalues of these primary matrices are their scalar values at the eigenvalues of $A$, so [F1] and [F2] give the stated strip and right-half-plane restrictions.

Suppose $Y$ is another logarithm of $A$ with spectrum in the open strip. On a Jordan block at $\mu \in \sigma (Y)$, $\exp (\mu )$ lies in $S$ and $\mathrm{Log}(\exp z)=z$ on a neighborhood of $\mu$. Step 1.1 therefore gives $\mathrm{Log}(\exp (J))=J$. Applying this on every block of $Y$ and using similarity invariance yields $L(A)=\mathrm{Log}(\exp (Y))=Y$.

Suppose $Q^2=A$ and $\sigma (Q)$ lies in the open right half-plane. Put $D=Q-R(A)$. Then $QD+DR(A)=Q(Q-R(A))+(Q-R(A))R(A)=Q^2-R(A{)}^2=0$. The spectra of $Q$ and $R(A)$ lie in the right half-plane, so $\sigma (Q)\cap (-\sigma (R(A)))=\mathrm{\varnothing }$. By [F4], the homogeneous Sylvester equation has only the zero solution; hence $D=0$ and $Q=R(A)$.