TITLE: Why does the least risky portfolio have large weight on the eigenvectors with small eigenvalues?
QUESTION [1 upvotes]: Suppose the average return of a portfolio P of N assets is defined as $R_p=\sum_{i=1}^Np_iR_i$, where $p_i \;(i=1,...,N)$ is the amount of capital invested in asset $i$, and $\{R_i\}$ are the expected returns of the individual assets. The risk on a portfolio can be associated to the total variance $\sigma^2_p=\sum_{i,j=1}^Np_iC_{ij}p_j$, where $C$ is the N by N covariance matrix of these assets.
The optimal portfolio, which minimizes $\sigma^2_p$ for a given value of $R_p$, can be found by Lagrange multiplier.
My question is, 

To solve this Lagrange multiplier, why does $C$ need to be inverted?
Why does eigensystem of C get involved?
Why does the least risky portfolio have large weight on the eigenvectors with small eigenvalues?

Thank you in advance.

REPLY [2 votes]: Let $\mathbf{R}$ denote  the vector of expected returns and $\mathbf{p}$ denote the vector of weights. 
The optimal portfolio is found by solving
$$\min_{\mathbf{P}}\sigma_p^2 = \min_{\mathbf{P}}\mathbf{p}^T\mathbf{C} \mathbf{p}$$
subject to the constraint
$$\mathbf{p}^T \mathbf{R} = R_p.$$
Introducing the Lagrange multiplier $\lambda$, we have the necessary first-order conditions ($1 \leqslant j \leqslant N):$
$$\frac{\partial}{\partial p_j}\left[ \mathbf{p}^T\mathbf{C} \mathbf{p} - \lambda (\mathbf{p}^T \mathbf{R}-R_p) \right] = 0 \\ \implies 2\mathbf{C} \mathbf{p} - \lambda \mathbf{R}= 0. $$
Thus, the optimal weights are given by
$$\mathbf{p}^* = \frac{\lambda}{2} \mathbf{C}^{-1}\mathbf{R}.$$
Apply the constraint to solve for $\lambda$.
(1) Now you can see why the inverse of the covariance matrix is relevant.
(2) Consideration of the eigensystem of $\mathbf{C}$ is not needed to find the optimal portfolio weights.
(3) Consider the fact the eigenvectors of the symmetric, positive definite matrix $\mathbf{C}$ form an orthonormal basis and expand $\mathbf{p}^*$ in this basis.
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