Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 1 / 7A (Vectors and Matrices)

Let $v\ne 0$ be an eigenvector of the Hermitian matrix $A$, with eigenvalue $\lambda$. Multiplying $Av=\lambda v$ by $v^{†}$ givesThe denominator is the positive real number $\Vert v{\Vert }^2$, whileThus the numerator is real, and every eigenvalue of a Hermitian matrix is real. This argument does not require distinct eigenvalues.

Take eigenvectors $v_i,v_j$ for distinct eigenvalues ${\lambda }_i,{\lambda }_j$. Since the eigenvalues are real and $A$ is a Hermitian matrix,Hence $({\lambda }_j-{\lambda }_i)v_i^{†}{v}_j=0$, so eigenvectors for distinct eigenvalues are orthogonal vectors. They are therefore linearly independent; after normalization the given $n$ eigenvectors form an orthonormal basis of ${\mathbb{C}}^n$.

Use the conventionfor the characteristic polynomial. Choosing $\det (A-tI)$ instead multiplies it by $(-1{)}^n$ and leaves the requested zero identity unchanged. For each eigenvector $v_j$, induction gives $A^r{v}_j={\lambda }_j^r{v}_j$, including $r=0$. Thusbecause ${\lambda }_{j}I-A$ has the nonzero kernel vector $v_j$ and therefore has zero determinant. Part (a) supplies a basis consisting of these $n$ eigenvectors. A linear map vanishing on every basis vector vanishes everywhere. ConsequentlyThis proves the Cayley-Hamilton theorem identity directly in the distinct-eigenvalue case assumed here.

Expand $x={\sum }_j{c}_j{v}_j$ in the orthonormal basis from part (a). ThenFor $x\ne 0$, the weights $|c_j{|}^2/{\sum }_k|c_k{|}^2$ are nonnegative and sum to one. The Rayleigh quotient therefore lies between the smallest and largest eigenvalues. Conversely, normalized extreme eigenvectors $v_{\min },v_{\max }$ give, for $0\le t\le 1$,Thus the exact Hermitian Rayleigh quotient range isWhen $n=1$, the range is the singleton containing the sole eigenvalue.

There is no such universal constant. For a real parameter $M$, consider the real matrixIts characteristic polynomial is $(t-1{)}^2$, so both eigenvalues are real and ${\lambda }_{\max }(B_M)=1$. ButFor any proposed finite $C$, choosing $M>2(C-1)$ violates the proposed bound. This is an explicit illustration that nonnormal eigenvalues do not bound a quadratic quotient.