Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / P (Numerical Analysis)

Gershgorin circle theorem places every complex eigenvalue of $A$ in at least one disk $D_i=\{z:|z-a_{ii}|\le {\sum }_{j\ne i}|a_{ij}|\}$. To prove it, take a nonzero eigenvector $v$ and choose $i$ with $|v_i|={\max }_j|v_j|>0$. The $i$th eigenvector equation givesDivision by $|v_i|$ proves the disk bound.

Here $C=I$, and the three disks for $B+\mu I$ have real centres $2+\mu$, $3+\mu$, $4+\mu$ and radii $2,3,4$ respectively. Every point in every disk has real part at least $\mu >0$. Hence zero cannot be an eigenvalue, proving $B+\mu C$ is nonsingular for every $\mu >0$.

For any $q\in {\Pi }_n$, the difference $p-q$ lies in ${\Pi }_n$ and is orthogonal to $f-p$. The Pythagorean theorem for inner product spaces givesThus the orthogonality condition proves minimality, and equality forces $q=p$.

Write the requested linear polynomial as $p=a+bx$. Orthogonality to one and $x$ givesOdd integrals vanish, leaving $2/5-2a/3=0$ and $-2b/5=0$. ConsequentlyThe weight's single zero does not destroy positive definiteness on continuous functions: a nonzero continuous function is nonzero on an interval containing points where $x^2>0$. The preceding projection argument therefore applies to this inner product.