Realification preserves Hermitian spectra, norms, and congruence

Identify $z=x+iy\in {\mathbb{C}}^n$ with $(x,y)\in {\mathbb{R}}^{2n}$. For any $Z=X+iY\in {\mathbb{C}}^{m\times n}$, direct multiplication gives $\mathcal{R}(Z)(x,y)=(\mathrm{\Re }(Zz),\mathrm{\Im }(Zz))$, so realification respects every compatible product. The displayed inner products make the standard coordinate bases orthonormal, so [F7] gives $Z^{\ast }=X^{\mathsf{T}}-i{Y}^{\mathsf{T}}$ and $\mathcal{R}(Z^{\ast })=\left(\begin{array}{cc}{X}^{\mathsf{T}}& Y^{\mathsf{T}}\\ -Y^{\mathsf{T}}& X^{\mathsf{T}}\end{array}\right)=\mathcal{R}(Z{)}^{\mathsf{T}}$.

For $z=x+iy$, $\mathrm{\parallel }z{\mathrm{\parallel }}_2^2={\sum }_j\mathrm{\mid }{x}_j+i{y}_j{\mathrm{\mid }}^2={\sum }_j(x_j^2+y_j^2)=\mathrm{\parallel }(x,y){\mathrm{\parallel }}_2^2$ by [F5, F6]. Thus for $n>0$ the identification bijects the complex and real unit spheres, so the maxima in [F4] coincide for $E$ and $\mathcal{R}(E)$. If $n=0$, both operator norms are zero by [F4]. If square $Z$ is invertible, $\mathcal{R}(Z^{-1})$ is an inverse of $\mathcal{R}(Z)$ by step 1.1; conversely, if $\mathcal{R}(Z)$ is invertible, its kernel is zero, so $Zz=0$ implies $z=0$, and [F3] makes $Z$ invertible.

If $n=0$, the spectral assertion is immediate. Otherwise, [F7] identifies the Hermitian matrix $H=H^{\ast }$ with a self-adjoint endomorphism in the standard orthonormal basis; [F2] makes it normal, so [F1] gives $H=U\mathrm{\Lambda }{U}^{\ast }$ with $U$ unitary and $\mathrm{\Lambda }=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$ real. By step 1.1, $\mathcal{R}(U)$ is real orthogonal and $\mathcal{R}(H)=\mathcal{R}(U)\mathcal{R}(\mathrm{\Lambda })\mathcal{R}(U{)}^{\mathsf{T}}$, where $\mathcal{R}(\mathrm{\Lambda })=\mathrm{diag}(\mathrm{\Lambda },\mathrm{\Lambda })$. Hence $\mathcal{R}(H)$ is real symmetric with each eigenvalue of $H$ repeated twice.