If $\sigma >0$ is a simple singular value with left and right singular vectors $u,v$, then its real directional derivative is $\mathrm{Re}(u^{\ast }Hv)$

Form the Hermitian block path $B(t):=\left(\begin{array}{cc}0& A+tH\\ (A+tH{)}^{\ast }& 0\end{array}\right)\mathrm{.}$ Then $w=\frac{1}{\sqrt{2}}(u,v{)}^{\mathsf{T}}$ is a unit eigenvector of $B(0)$ with eigenvalue $\sigma$, because $B(0)w=\frac{1}{\sqrt{2}}(Av,A^{\ast }u{)}^{\mathsf{T}}=\frac{1}{\sqrt{2}}(\sigma u,\sigma v{)}^{\mathsf{T}}=\sigma w\mathrm{.}$ If $B(0)(x,y{)}^{\mathsf{T}}=\sigma (x,y{)}^{\mathsf{T}}$, then $Ay=\sigma x$ and $A^{\ast }x=\sigma y$, so $A^{\ast }Ay={\sigma }^{2}y$. Since $\sigma$ is a simple positive singular value, the eigenspace of $A^{\ast }A$ for ${\sigma }^2$ is one-dimensional, and then $x={\sigma }^{-1}Ay$ is determined by $y$. Hence $\sigma$ is a simple eigenvalue of the Hermitian matrix $B(0)$.

Because $t\mapsto B(t)$ is differentiable and step 1.1 shows that $\sigma$ is a simple eigenvalue of $B(0)$, [L2] gives a local $C^1$ eigenvalue branch through $\sigma$. The derivative of the block path is $B^{\mathrm{\prime }}(0)=\left(\begin{array}{cc}0& H\\ H^{\ast }& 0\end{array}\right)\mathrm{.}$ Applying [L1] to this Hermitian simple eigenvalue branch gives $D\sigma (A)[H]=w^{\ast }{B}^{\mathrm{\prime }}(0)w\mathrm{.}$

Expanding the quadratic form from step 2.1 gives $D\sigma (A)[H]=\frac{1}{2}(u^{\ast }Hv+v^{\ast }{H}^{\ast }u)=\mathrm{Re}(u^{\ast }Hv)\mathrm{.}$