Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 1 / 7B (Vectors and Matrices)

Let $Av=\lambda v$ for a nonzero eigenvector $v$ of the Hermitian matrix $A$. Hermitian symmetry of the inner product givesSince $lVertv{\Vert }^2>0$, $lambda=\overline{\lambda }$, so every eigenvalue is real.

For an orthogonal projection matrix $P=P^{†}=P^2$,Thus $P$ is a positive semidefinite matrix.

SetThen $v=v_0+v_1$, and idempotence givesMoreover,because both $P$ and $I-P$ are Hermitian. This is the orthogonal decomposition into the kernel and image of the projection.

Put $H=A-B$. This is a nonzero Hermitian matrix. By the finite-dimensional spectral theorem, it has a unit eigenvector $v$ with a nonzero real eigenvalue $lambda$. Define the rank-one orthogonal projection matrixThe cyclic property of the matrix trace givesHence this $P$ distinguishes $A$ and $B$ through the required traces.

No. Consider the two orthogonal projectionsTheir product isFor $x=(-1,2{)}^T$,Thus $PQ$ is not a positive semidefinite matrix. It is also not Hermitian; products of orthogonal projections need not remain orthogonal projections unless the factors commute.