You will have to add some constraints to get the weight vector of the eigen vector of the smallest eigen values, otherwise 0 is a trivial solution.

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. 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.

The way to minimize a positive linear combination of positive terms is to allocate the minimum amount of weight possible to the smallest term. As soon as you start to allocate some weight to a bigger term, you will have a bigger number.

You will have to add some constraints to get the weight vector of the eigen vector of the smallest eigen values, otherwise 0 is a trivial solution.

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.

The way to minimize a positive linear combination of positive terms is to allocate the minimum amount of weight possible to the smallest term. As soon as you start to allocate some weight to a bigger term, you will have a bigger number.

You will have to add some constraints to get the weight vector of the eigen vector of the smallest eigen values, otherwise 0 is a trivial solution.

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. 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.

The way to minimize a positive linear combination of positive terms is to allocate the minimum amount of weight possible to the smallest term. As soon as you start to allocate some weight to a bigger term, you will have a bigger number.

You will have to add some constraints to get the weight vector of the eigen vector of the smallest eigen values, otherwise 0 is a trivial solution.

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. 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.

The way to minimize a positive linear combination of positive terms is to allocate the minimum amount of weight possible to the smallest term. As soon as you start to allocate some weight to a bigger term, you will have a bigger number.