Let $X$ be a $p$-dimensional random vector with $p$ principal components $y_1, y_2, \dots, y_p$. By definition, a restriction put on the second principal component $y_2 = a_2'X$ is

$$ \text{cov}(y_1, y_2) = \text{cov}(a_1' X, a_2' X) = 0 $$

It is also known that $a_i$ are simply eigenvectors, so we also have

$$ \forall i \ne j, a_i \cdot a_j = 0 $$

I wonder if it's a coincidence in the context of PCA. In other words, does $\text{cov}(a_1' X, a_2' X) = 0$ imply $a_1 \cdot a_2 = 0$? I tried to prove it, but didn't get very far

$$ \begin{equation} \begin{aligned} 0 &= \text{cov}(a_1' X, a_2' X) \\ &= \text{cov}(\sum_{k=1}^p a_{1,k} X_k, \sum_{m=1}^p a_{2,m} X_m) \\ &= \sum_{k=1}^p a_{1,k} \sum_{m=1}^p a_{2,m} \text{cov}(X_k, X_m) \end{aligned} \end{equation} $$

Can I get $a_1 \cdot a_2 = 0$ from here?

  • 2
    $\begingroup$ Geometrically, you are asking whether two vectors remain orthogonal after an arbitrary linear transformation. In light of this it's easy to find counterexamples in just $p=2$ dimensions. In fact, you are almost certain to find a counterexample just by randomly selecting any $2\times 2$ covariance matrix. $\endgroup$
    – whuber
    May 12, 2019 at 16:08

1 Answer 1


Consider a random walk model, $X_t = X_{t-1} + Z_t$ where $Z_t \overset{iid}{\sim} \text{Normal}(0,1)$. Also, assume that $X_1 = Z_1$ and has variance $1$. Take the random vector $X = (X_1,X_2,X_3)'$, it has a covariance matrix $$ \Sigma = \left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array}\right]. $$ Now take $a_1 = (-1,1,0)'$ and $a_2 = (0,-1,1)$. You can see that $$ \text{Cov}(a_1'X, a_2'X) = \text{cov}(Z_2, Z_3) = 0 $$ but $a_1'a_2 = -1$.

Edit: just for completeness, consider the example that your mind might go to first. Consider a vector $X$ such that its covariance matrix is proportional to the identity matrix (i.e. $\sigma^2 \mathbf{I}$ with $\sigma^2 >0$). Then $\text{Cov}(a_1'X, a_2'X) = \sigma^2 a_1' a_2$ is zero if and only if $a_1' a_2 = 0$.


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