They should sum to $100~\%.$ The total variance of a $p$-variate random variable $X$ with covariance matrix $\Sigma$ is defined as $${\rm tr}(\Sigma)=\sigma_{11}+\sigma_{22}+\cdots+\sigma_{pp}.$$ Now, the trace of a symmetric matrix is the sum of its eigenvalues $\lambda_1\geq\lambda_2\geq\ldots\geq\lambda_p.$ Thus the total variance is $${\rm tr}(\Sigma)=\lambda_1+\cdots+\lambda_p$$ if we use $\lambda_i$ to denote the