Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 6 / i

The decomposition is $A=H+iS$. Both matrices are Hermitian: $H^{†}=H$ and $S^{†}=S$, directly from their definitions and complex conjugation of $1/(2i)$. The matrix $H$ is the Hermitian part of a matrix, whereas $S$ is the Hermitian matrix obtained by dividing the skew-Hermitian part by $i$.

Multiply the eigenvalue equation $Ax=\lambda x$ on the left by $x^{†}$. The unit vector normalization is $x^{†}x=1$, soNo normal matrix assumption on $A$ is needed: this is an identity for each normalized eigenvector.

For any Hermitian matrix $K$, the scalar $x^{†}Kx$ is real because its complex conjugate equals $x^{†}{K}^{†}x=x^{†}Kx$. Applying this to $H,S$ in part (a) givesThese are the real and imaginary quadratic forms associated with the numerical range of $A$.