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

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 $A$.

Use the inner product $⟨u,v⟩=u^{†}v$ on ${\mathbb{C}}^n$. The distinct real eigenvalues permit a full orthonormal eigenvector expansion; the two child solutions establish that expansion and apply it to the proposed quadratic form.

For $B{x}_i={\lambda }_i{x}_i$, Hermitian symmetry givesDistinct eigenvalues therefore imply $x_i^{†}{x}_j=0$ for $i\ne j$. Together with the unit normalizations,The $n$ mutually orthogonal nonzero vectors are linearly independent and span the $n$-dimensional space, so they form an orthonormal basis. Here ${\delta }_{ij}$ is the Kronecker delta.

Expand $v={\sum }_j{c}_j{x}_j$, where $c_j=x_j^{†}v$. By orthonormality,Since the spectrum avoids $[a,b]$, each ${\lambda }_j$ is either below $a$ or above $b$, making every product strictly positive. This proves positivity of a quadratic polynomial across a spectral gap:In particular $\beta >0$ for every nonzero $v$. For $v=0$, $\beta =0$; the printed strict claim for an arbitrary vector must exclude this case.