# Markowitz Eigenvalues

I came across this passage in a book about PCA and denoising of Markowitz:

But eigenvalues that are important from risk perspective are least important ones from portfolio optimization perspective. Information matrix mixes the returns and has Reciprocal values 1/λ k of the eigenvalues of the covariance matrix λ . This is one reason why portfolio managers often do not use portfolio optimization methods.

What I understand is that the eigenvalues of the covariance matrix contribute the most to the risk in the portfolio (i.e. large eigenvalues correspond to asset with high variances and covariances). But why is that problematic in the information matrix, which mixes the returns in the Markowitz solution?

I mean why would we not consider it a good thing, if the largest eigenvalues of the covariance_matrix become the smallest eigenvalue in the information matrix? From my understanding, this would mean that we don't give them too much weight and we therefore avoid heavy correlations? But it's obviously the other way round...

The main problem stems from the case opposite of the one that you are focusing on: The inversion of the covariance matrix leads to a situation where the smallest eigenvalues of the covariance matrix (representing the least significant sources of market risk) become the largest eigenvalues in the information matrix (because $$1/\lambda_{\text{small}}$$ is large). This means these lesser risk components get the most weight in the optimization, and as a result, they can have an undue influence on the portfolio allocation.
This can be seen more clearly through the following derivation. The basic Markowitz problem we are trying to solve is given by $$\begin{equation} x(t) = \underset{x \in \mathbb{R}^n}{\text{argmax }} x^T\mu \end{equation}$$ \begin{align} \text{s.t. } x^TQx \leq \sigma_{\text{max}}^2 \end{align} The analytical solution is given by $$x^* = \sigma_{\text{max}}\frac{Q^{-1}\mu}{\mu^T Q^{-1}\mu}.$$ The inverse of $$Q$$ has the eigendecomposition $$Q^{-1} = V D^{-1} V^T,$$ where $$V$$ is the matrix of eigenvectors and $$D$$ the diagonal matrix with eigenvalues. The decomposition can be presented as an outer product in the following way: $$Q^{-1} = \sum_{i=1}^{n} \frac{1}{\lambda_i} v_i v_i^T.$$ Plugging this back into the analytic formula yields $$x(t) = \frac{\sigma_{\text{max}}}{\mu^T Q^{-1} \mu} \sum_{i=1}^{n} \frac{v_i^T \mu}{\lambda_i} v_i.$$ Thus we can see that the optimal solution to the standard Markowitz problem is most heavily influenced by the the eigenvectors with the smallest eigenvalue of the covariance matrix. Especially if $$\lambda_i$$ is extremely close to zero, the corresponding eigenvector will dominate all the others, causing the resulting portfolio to "blow up".