Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4 / 40E / a / Solution

Let $Mx=(A-B)x$ and suppose $Hx=\lambda x$, so $-Bx=\lambda Mx$. Put $a=x^{\ast }Ax>0$ and $m=x^{\ast }Mx$. Since $A=M-(-B)$, $a=(1-\lambda )m$. AlsoPositive definiteness makes this positive, hence $|\lambda |<1$ for every eigenvalue and $\rho (H)<1$.

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