Let $A$ be a real symmetric $n \times n$ matrix.

(a) Prove the following:

(i) Each eigenvalue of $A$ is real and there is a corresponding real eigenvector.

(ii) Eigenvectors corresponding to different eigenvalues are orthogonal.

(iii) If there are $n$ distinct eigenvalues then the matrix is diagonalisable.

Assuming that $A$ has $n$ distinct eigenvalues, explain briefly how to choose (up to an arbitrary scalar factor) the vector $v$ such that $\frac{v^{T} A v}{v^{T} v}$ is maximised.

(b) A scalar $\lambda$ and a non-zero vector $v$ such that

$$A v=\lambda B v$$

are called, for a specified $n \times n$ matrix $B$, respectively a generalised eigenvalue and a generalised eigenvector of $A$.

Assume the matrix $B$ is real, symmetric and positive definite (i.e. $\left(u^{*}\right)^{T} B u>0$ for all non-zero complex vectors $u$ ).

Prove the following:

(i) If $\lambda$ is a generalised eigenvalue of $A$ then it is a root of $\operatorname{det}(A-\lambda B)=0$.

(ii) Each generalised eigenvalue of $A$ is real and there is a corresponding real generalised eigenvector.

(iii) Two generalised eigenvectors $u, v$, corresponding to different generalised eigenvalues, are orthogonal in the sense that $u^{T} B v=0$.

(c) Find, up to an arbitrary scalar factor, the vector $v$ such that the value of $F(v)=\frac{v^{T} A v}{v^{T} B v}$ is maximised, and the corresponding value of $F(v)$, where

$$A=\left(\begin{array}{ccc} 4 & 2 & 0 \\ 2 & 3 & 0 \\ 0 & 0 & 10 \end{array}\right) \quad \text { and } \quad B=\left(\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 3 \end{array}\right)$$