Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 4

A Hermitian matrix obeys $H^{†}=H$, where the dagger is conjugate transpose. If $Hv=\lambda v$ with $v\ne 0$, then $v^{†}Hv=\lambda v^{†}v$. This scalar is real because its complex conjugate equals $v^{†}{H}^{†}v=v^{†}Hv$. Since $v^{†}v>0$, the eigenvalue $\lambda$ is real.

For the Gram matrix, conjugate symmetry of the inner product givesThus $G$ is Hermitian. The assumption that the vectors form a basis additionally makes $G$ positive definite, though only Hermiticity is needed here.

Let $A{v}_1={\lambda }_1{v}_1$ and $A{v}_2={\lambda }_2{v}_2$, with both vectors nonzero and ${\lambda }_1\ne {\lambda }_2$. If $c_1{v}_1+c_2{v}_2=0$, apply $A-{\lambda }_{2}I$ to get $c_1({\lambda }_1-{\lambda }_2)v_1=0$. Hence $c_1=0$, and then $c_2=0$.

For the general independence of eigenvectors for distinct eigenvalues, use induction. In a relation ${\sum }_{j=1}^n{c}_j{v}_j=0$, applying $A-{\lambda }_{n}I$ gives ${\sum }_{j=1}^{n-1}{c}_j({\lambda }_j-{\lambda }_n)v_j=0$. Independence of the first $n-1$ vectors makes each coefficient vanish. Distinctness of the eigenvalues then gives $c_1=\cdots =c_{n-1}=0$, and the original relation gives $c_n=0$. No Hermitian assumption is required.