# Proof for principal component score

Can we proof that two principal component scores are not equal or one is better than the other in terms of variation? For example, the 1st principal component score is defined as $z_1 = a_{11}(y_1 - \mu_1) +...+ a_{1p}(y_p - \mu_p)$ with the eigenvalue $\lambda_1$. We all know that the proportion of total population variance due to $j^{th}$principal component is determined by its eigenvalue, such as $w_j = \frac{\lambda_j}{\lambda_1 + ... +\lambda_p}$. If we calculate a new score such as $s = z_1\times w_1 + ... + z_p\times w_p$. In addition, we also know that,

Var$(z_1) = \lambda_1$, Cov$(z_i, z_j) = 0$, total population variance = $\lambda_1 + ... + \lambda_p$, and $\Sigma{w_p} =1$.

My question is can we proof that Var$(s) = \lambda_1\times {w_1}^2 + ... + \lambda_p\times {w_p}^2 \geq$ or $\leq \lambda_1$?

${\rm Var}(s)\leq \lambda_1$. To see this, note that we have $${\rm Var}(s)=\frac{\sum_{i=1}^p \lambda_i^3}{(\lambda_1+\cdots+\lambda_p)^2}=\lambda_1\cdot\frac{\lambda_1^2+\sum_{i=2}^p\frac{\lambda_i^3}{\lambda_1}}{(\lambda_1+\cdots+\lambda_p)^2}.$$ But $$(\lambda_1+\cdots+\lambda_p)^2\geq \lambda_1^2+\cdots+\lambda_p^2\geq \lambda_1^2+\sum_{i=2}^p\frac{\lambda_i^3}{\lambda_1}$$ since $\lambda_1\geq\lambda_i$ for $i=2,3,\ldots,p$.
Thus $$\frac{\lambda_1^2+\sum_{i=2}^p\frac{\lambda_i^3}{\lambda_1}}{(\lambda_1+\cdots+\lambda_p)^2}\leq 1$$ and $${\rm Var}(s)=\lambda_1\cdot\frac{\lambda_1^2+\sum_{i=2}^p\frac{\lambda_i^3}{\lambda_1}}{(\lambda_1+\cdots+\lambda_p)^2}\leq \lambda_1\cdot 1=\lambda_1.$$