# Why the weight vector of 'global minimum variance' the 'eigenvector' with the minimum eigenvalue?

## Question

Why is it the case that the weight vector of the global minimum variance portfolio the eigenvector of the covariance matrix with the smallest eigenvalue?

## Question with more details

1. I know the relationship between PCA and eigenvector. If we want to compress data in 1 dimension, we should use Lagrangian with covariance matrix of the sampled data.
2. By doing so, we can find the a vector that can produce maximum amount of varinace of data points when they are projected onto the vector. This vector is called the 1st principal component with the greatest eigenvalue.
3. You can read more about the meaning of eigenvectors in a covarinace matrix in this blog post
4. However, in global minimum variance portfolio, what we want to do is more of reverse to PCA. We want to find the vector that produces the least amount of variance when data points are projected onto the line. As such, I don't think we can use Lagrangian.
5. However this blog post still argues that the weight vector in global minimum variance is the eigenvector of the covariance matrix with the smallest eigenvalue. Can anyone tell me why, please?

Without going in the details of handling those extra constraints, the reason why the vector space associated with the smallest eigen value is relevant is because if you express variance of your portfolio in the eigen basis, you have $$\sigma^2=\Sigma_i{\sigma_i^2 \omega_i^2}$$ with $$\omega_i$$ beeing the coordinates of your portfolio in the eigen space of the covariance matrix.
The proof of that is by direct application of the definition of what an eigen basis is. If W is your weight vector in the canonical basis, and $$\omega$$ the weight vector in the eigen basis. By definition of the eigen basis, you have the covariance matrix $$M=P'SP$$ with $$S$$ a diagonal matrix of coefficients $$\sigma_i^2$$ and $$P$$ the transformation matrix to go from the canonical basis to the eigen basis. ($$P'$$ is $$P$$ tranposed) i.e. $$\omega=PW$$. Hence you have:$$\sigma^2=W'MW=W'P'SPW=\omega'S\omega=\Sigma_i\sigma_i^2\omega_i^2$$
You can see that if you try to minimize this variance with $$\omega$$ unknown, you have to minimize a sum of positive terms with positive coefficients. Hence the minimum is reached when all are $$\omega_i=0$$, if not possible, then you will allocate some weight to the smallest number possible and none everywhere else.