Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 5 / b

The conjugate transpose reverses the order of matrix multiplication. Since $A$ and $B$ are Hermitian matrices,Consequently $\boxed{AB\text{ is Hermitian}⟺AB=BA}$: the condition is exactly that the two matrices commute.

Take $\boxed{c=1}$. Indeed the conjugate transpose isso the anticommutator is a Hermitian matrix. The original PDF, p. 6, supplies the factor $BA$ that is corrupted in the local OCR.

Using the conjugate transpose and the Hermitian matrix assumptions,Thus $ABA$ is a Hermitian matrix, whether or not $A$ and $B$ commute.

Put $D=A-iB$, so $D^{†}=A+iB$. Because the Hermitian matrices $A,B$ commute,For every nonzero vector $v$,since $A$ is an invertible matrix. Thus $D$ and $D^{†}$ are invertible matrices, justifying the inverse in the definition of $U$. The conjugate transpose of $U=D^{†}{D}^{-1}$ is $(D^{†}{)}^{-1}D$. Since $D$ and $D^{†}$ commute, their inverses do too, andHence $U$ is a unitary matrix.

For the unitary matrix $C$, $C^{-1}=C^{†}$. Therefore $D=C^{-1}AC$ is a Hermitian matrix:Every positive integer power remains a Hermitian matrix, since $(D^p{)}^{†}=(D^{†}{)}^p=D^p$. Its eigenvalues are real by the finite-dimensional spectral theorem. Equivalently, matrix similarity giveswhose eigenvalues are $\boxed{{\lambda }_j^p\in \mathbb{R}}$, where the ${\lambda }_j$ are the real eigenvalues of $A$.

Complete the square in the Hermitian matrix polynomial:For every nonzero vector $v$,Thus $H$ is a positive-definite matrix, so $\boxed{0\text{ is not an eigenvalue of }H}$. In finite-dimensional spectral theorem terms, each eigenvalue isIf ${\alpha }^2\ge 4\beta$, the scalar polynomial has real roots of a polynomialAn allowed nonzero root can be an eigenvalue of $A$, a Hermitian matrix which is also an invertible matrix, in which case $H$ has a zero eigenvalue. For example, $\alpha =2$, $\beta =1$, $A=-I$ gives $H=0$. The weaker inequality does not force failure for every $A$; it removes the general positive lower bound. There is also the harmless corner $\alpha =\beta =0$, for which $H=A^2$ stays positive definite because $A$ is invertible, even though the strict discriminant hypothesis is absent.