Can PC1 explain more than 90% of variance? I've been running some analysis using morphological data. However, when I run a PCA, PC1 explains more than 90% of variance. I'm not sure if this is possible. If not, what should I need to do?


Let $S \in \mathbb{R}^{p \times p}$ be the sample covariance matrix (which equals to $X'X$ for centered and intercept-free design matrix $X$), and $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_p \geq 0$ be all the non-negative eigenvalues of $S$, then the total variation percentage explained by the first $k$ PCs is given by
\begin{align}
p_k = \frac{\sum_{j = 1}^k \lambda_j}{\sum_{j = 1}^p \lambda_j}, \; k = 1, 2, \ldots, p.
\end{align}
In particular, the total variation percentage explained by the first PC is
\begin{align}
p_1 = \frac{\lambda_1}{\sum_{j = 1}^p\lambda_j}. \tag{1}
\end{align}
In view of $(1)$, theoretically there are no upper bounds that are strictly less than $1$ for $p_1$.  In other words, there exists $X$ such that $p_1$ can be arbitrarily close to $1$. For example, any design matrix $X$ in the form of $(2)$ below satisfies that its first PC explains $1 - \varepsilon$ total variation:
\begin{align}
X = U\begin{bmatrix}
\operatorname{diag}(\sqrt{1 - \varepsilon}, c, \ldots, c) \\
0 
\end{bmatrix}V', \tag{2}
\end{align}
where $c = \frac{\sqrt{\varepsilon}}{\sqrt{p - 1}}$, $U$ and $V$ are arbitrary order $n$ and order $p$ orthogonal matrices. Therefore, it is not surprising at all to observe cases such that $p_1 > 90\%$.
