Assume we have a family of $$n$$ assets whose daily return is vector $$\mu$$ and the covariance matrix is $$\Sigma$$. If we make a portfolio whose asset weight is $$w_k$$ for asset $$k$$ and the vector of all weights is $$w$$, then the portfolio’s volatility is $$\sigma = (w^\top\Sigma w)^{1/2}$$.

Now, for matrix $$\Sigma$$, we can have spectral decomposition

$\Sigma V = V \Lambda$

where $$V$$ is matrix of columns of eigenvectors and $$\Lambda$$ is a diagonal matrix of eigenvalues (in descending order), both of $$\Sigma$$. Hence we can write $$\Sigma = V \Lambda V^\top$$. Substitute this into the formula for the square of volatility, we have

\begin{aligned} \sigma^2 &= w^\top \Sigma w \\ &= w^\top (V\Lambda V^\top) w \\ &= (w^\top V) \Lambda (V^\top w) \end{aligned}

if we substitute $$u = V^\top w$$, we have

\begin{aligned} \sigma^2 &= (w^\top V) \Lambda (V^\top w) \\ &= u^\top \Lambda u \end{aligned}

Now since $$\Lambda$$ is diagonal,

$\Lambda = \text{diag}\begin{pmatrix}\lambda_1 & \lambda_2 & \dots & \lambda_n\end{pmatrix}$

with $$\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_n$$, therefore

$\sigma^2 = \sum_{k=1}^n u_k^2\lambda_k$

This is significant because normally the variance $$\sigma^2$$ is not additive but we can make it so by looking at the eigenvector dimensions. Also, if $$\Sigma$$ is positive definite (usually the case for covariance matrix) then $$\lambda_k > 0$$ for all $$k$$. This is a useful information on the problem of finding the minimal $$\sigma^2$$ subject to some constraint on $$u$$. Assume we have the condition that $$\sum_k u_k=C$$ for some constant $$C>0$$, it is trivial to see that the way to minimize $$\sigma^2$$ is to make $$u_k=0$$ for $$k=1,\dots,n-1$$ and $$u_n=C$$. In this case, the portfolio volatility is $$\sigma = u_n\sqrt{\lambda_n} = C\sqrt{\lambda_n}$$. In this case, $$C$$ is just a scaling factor for the size of the portfolio. If we have $$u$$, it is easy to reverse it to get

$w = V u.$

Since $$V$$ and $$\Lambda$$ are the eigenvector and eigenvalues of $$\Sigma$$, we can consider the mutually orthogonal eigenvectors are different “risk dimensions” and a portfolio is constructed so as to distribute the total risk to these different dimensions. In this case, we can write $$R=\begin{pmatrix}r_1 & r_2 & \dots & r_n\end{pmatrix}$$ as a vector of risk proportions to each dimension and the risk distribution satisfies

\begin{aligned} r_k &= \frac{u_k^2 \lambda_k}{\sigma^2} \\ u_k &= \sqrt{\frac{r_k\sigma^2}{\lambda_k}} = \sigma\sqrt{\frac{r_k}{\lambda_k}} \end{aligned}

Hence we can construct a portfolio with arbitrary allocation of risk to each dimensions.