Hi. 


I am busy working through a paper i came accross online on portfolio optimization.
The paper may be accessed on the following link:
<http://ssrn.com/abstract=1483412>
I am struggling, in particular, with the equation 5 on page 4. I am not sure how the authors managed to derive the equation. Why are only the first and last eigenvalues used??? 


Any ideas would be great!!!!


Many thanks in advance.



Relating the angle between two vectors to max and min eigenvalues
Problem: Let $T$ be a positive definite selfadjoint operator in an $n$-dimensional inner product space $H$. Find the maximum possible angle between a vector $v\neq 0$ and its image $Tv$, expressed in terms of the eigenvalues $\theta\_0^2\geq\dots\geq\theta\_{n-1}^2>0$ of $T$.


Solution: 
We want to minimize $\cos(v,Tv)=\frac{\langle v,Tv \rangle}{\|v\|\|Tv\|}$, or equivalently, its square $\frac{\langle v,Tv\rangle^2}{\langle v,v\rangle\langle Tv,Tv\rangle}$.


Since the angle between any vector $v\neq 0$ and its image $Tv$ doesn't change if we rescale $v$, we can rescale at our wish. I choose to restrict to those $v$ such that the numerator $g(v)=\langle v,Tv\rangle$ equals 1, so now the problem is equivalent to minimizing the denominator $f(v)=\langle v,v\rangle\langle Tv,Tv\rangle$, and now there are no divisions bothering.


Critical points of this restricted function are given by the equation $df(v)=\lambda dg(v)$, where $\lambda\in\mathbb R$ is a Lagrange multiplier. We calculate


$dg(v)=\langle v,T-\rangle+\langle -,Tv\rangle=2\langle Tv,-\rangle$ and


$df(v)=2\langle v,-\rangle\langle Tv,v\rangle+2\langle v,v\rangle\langle Tv,T-\rangle=2\|Tv\|^2\langle v,-\rangle+2\|v\|^2\langle T^2v,-\rangle$. 


The critical point equation can now be rewritten:


$\langle \|Tv\|^2v+\|v\|^2T^2v-\lambda Tv,-\rangle=0$


and this is true iff $\|Tv\|^2v+\|v\|^2T^2v-\lambda Tv=0$. So $v$, $Tv$, and $T^2v$ are linearly dependent when $v$ is a critical point. But we know by Vandermonde [...] that they would be independent if $v$ had nonzero projections in three eigenspaces. 


So the critical points are found in the planes spanned by two eigenvectors, and then we must solve our problem for $n=2$, being only interesting the case in which the two eigenvalues are different, because otherwise $v$ will also be an eigenvector and then the angle will be zero. 


Calculations seem to get a little nicer if we express $T=S^2$, by letting $S$ be the only positive selfadjoint square root of $T$, with eigenvalues $\theta\_0\geq\dots\theta\_{n-1}>0$. We then have $g(v)=\langle Sv,Sv\rangle=\|Sv\|^2$, and $f(v)=\|v\|^2\|S^2v\|^2$.


Also, the symmetry is greater if we work in terms of the variable $w=Sv$, so that $g=\|w\|^2$ and $f=\|S^{-1}w\|^2\|Sw\|^2$ (this is not very important). If we express $w=x v\_i+ y v\_j$, where $v\_i$ and $v\_j$ are eigenvectors of $S$ with eigenvalues $\theta\_i$ and $\theta\_j$ and $i>j$, the Lagrange multipliers equation can be solved after some calculations [...], finding four critical points $(x,y)=(\pm \sqrt{\frac 12},\pm \sqrt{\frac 12})$. The minimum cosine for $w$ in the plane spanned by $v\_i$ and $v\_j$ is then $\frac{\langle w,w\rangle}{\|S^-1w\|\|Sw\|}=\dots=\frac{\theta\_i\theta\_j}{2(\theta\_i^2\theta\_j^2)}=\frac 12 (\frac{\theta\_i}{\theta\_j}+\frac{\theta\_j}{\theta\_i})$.


Once the case $n=2$ is solved, we want to select the plane so that $(\frac{\theta\_i}{\theta\_j}+\frac{\theta\_j}{\theta\_i})$ is maximum. But the function $h(t)=t+t^{-1}$ increases in the interval $[1,+\infty)$, so the value $t=\frac{\theta\_i}{\theta\_j}$ should be chosen as large as possible by maximizing $\theta\_i$ and minimizing $\theta\_j$.


The dots "..." represent explanations that I'm omitting. Request further details as needed.


ADDED by request of the OP: The "Vandermonde principle" says that an $n\times n$ [Vandermonde matrix](http://en.wikipedia.org/wiki/Vandermonde_matrix) with different rows is invertible. This implies that if a vector $v$ has nonzero components in the eigenspaces corresponding to three different eigenvalues $\lambda\_0$, $\lambda\_1$, $\lambda\_2$, then $v$, $Tv$ and $T^2v$ are independent. 


Proof: Write $v=v\_0+v\_1+v\_2$ so that for each $0\leq i<3$ we have $v\_i$ a nonzero eigenvector corresponding to eigenvalue $\lambda\_i$. Then we also have


$Tv=\lambda\_0v\_0+\lambda\_1v\_1+\lambda\_2v\_2$ and


$T^2v=\lambda\_0^2+\lambda\_1^2v\_2+\lambda\_2^2v\_2$.


Our vectors $v$, $Tv$ and $T^2v$ belong to the subspace $S$ with basis $B=(v\_0,v\_1,v\_2)$. And the coordinates of $v$, $Tv$ and $T^2v$ in basis $B$ are


$[v]\_B=\left(\begin{array}{c}1\\1\\1\end{array}\right)$


$[Tv]\_B=\left(\begin{array}{c}\lambda\_0\\ \lambda\_1\\ \lambda\_2\end{array}\right)$


$[T^2v]\_B=\left(\begin{array}{c}\lambda\_0^2\\ \lambda\_1^2\\ \lambda\_2^2\end{array}\right)$


These are the columns of a Vandermonde matrix, so they are linearly independent, and so must be the vectors $v$, $Tv$, $T^2v$.


