Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 1E (Linear Algebra)

For $n\ge 1$, an eigenvalue is a number $\lambda \in \mathbb{C}$ for which there is a nonzero vector $v$ satisfying $Av=\lambda v$. Equivalently, $\ker (A-\lambda I)\ne \{0\}$, or $\det (A-\lambda I)=0$. The characteristic polynomial $\det (tI-A)$ has degree $n$ and leading coefficient one. By the fundamental theorem of algebra it has a complex root $\lambda$. At that root $A-\lambda I$ is a singular matrix, so it has a nonzero kernel vector, proving that $A$ has an eigenvalue.

True. If $Av=\lambda v$ with $v\ne 0$ and $A=A^{\ast }$, thenSince $v^{\ast }v>0$, $\lambda =\bar{\lambda }$. Thus every eigenvalue of a Hermitian matrix is real.

False. The matrix $A=\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)$ has real eigenvalues $1$ and $2$, because its characteristic polynomial is $(t-1)(t-2)$, but $A^{\ast }=\left(\begin{array}{cc}1& 0\\ 1& 2\end{array}\right)\ne A$. Real spectrum alone does not imply that it is a Hermitian matrix.

False. Every entry of $A=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$ is strictly positive, yet $A(1,-1{)}^T=-(1,-1{)}^T$. Its eigenvalues are $3$ and $-1$, so one has negative real part. Positivity of entries does not imply the positive-definite matrix property.

False in general. Take $A=\mathrm{diag}(1,6,6)$ and $B=\mathrm{diag}(2,2,9)$. Both have trace $13$ and determinant $36$, but their eigenvalue multisets are different. For $n>3$, append equal identity blocks to preserve this counterexample. In dimensions one and two the assertion is true: in dimension two the characteristic polynomial is $t^2-(\mathrm{tr}A)t+\det A$, whose roots, with multiplicity, are determined by those two quantities.