Suppose that the $p$-dimensional random vector $X = [X1,\dotsc,Xp]^T$ has mean $\mathbb{E}(X) = \mu$ and positive definite covariance matrix $\text{Cov}(X) = \Sigma = {\sigma_{ij}}$. Also suppose that $\Sigma$ has distinct eigenvalues $\lambda_1,\dotsc, \lambda_p$ where $\lambda_1 >\dotsb> \lambda_p > 0.$ Let $e_1,\dotsc,e_p$ denote the eigenvectors of unitary length corresponding to the eigenvalues $\lambda_1,..., \lambda_p,$ respectively. Also let the $j$’th element of $e_i$ be denoted $e_{ij}$.

Let $X = [X_1,\dotsc,X_p]^T$ be a random vector and let $c = [c_1,\dotsc,c_p]^T$ be a vector of constants.

  1. The linear combination $c^T X = c_1 X_1 +\dotsb+ c_p X_p$ has $\mathbb{E}(c^T X) = c^T \mathbb{E}(X)$ and $\text{Var}(c^T X) = c^T \text{Cov}(X) c.$
  2. $\text{Cov}(a^T X, c^T X) = a^T \text{Cov}(X) c.$

Using the above show that, for a constant vector a, $\text{Cov}(a^TX, e^T_i(X − µ)= \lambda_ia^Te_i.$

I'm just not sure how to prove this.
I know $\text{Var}(a^TX) = a^T\Sigma a$ and $\text{Var}(e^T_i(X − \mu)) = \lambda_i.$

Any suggestions?

  • $\begingroup$ Is this different from your recent post "Show that, for a constant vector a, $Cov(a^TX, e_i^T(X- µ)) = λ_i a^Te_i$?"? Looks pretty much the same to me. $\endgroup$ Commented Sep 7, 2016 at 8:01
  • $\begingroup$ Yeah the same, I didn't think I explained it properly.I've deleted it now $\endgroup$
    – Stephen .S
    Commented Sep 7, 2016 at 8:03
  • $\begingroup$ Please delete one of the posts then. Here we do not post things twice. If we need to edit the original post, we do exactly that (there is an edit button below) rather than posting anew. $\endgroup$ Commented Sep 7, 2016 at 8:43

1 Answer 1


Because of $\text{Cov}$ linearity you can write:


I am barely using your second formula and the definition of $\text{Cov}$:


I can move out of the expectation all fixed quantities and $\text{Cov}(X)=\Sigma$:

$$\text{Cov}(a^{T}X,e_{i}^{T}(X-\mu))=a^{T}\Sigma e_{i}-a^{T}\mathbb{E}(X)e_{i}^{T}\mu+a^{T}\mathbb{E}(X)e_{i}^{T}\mu.$$

By definition of an eigen vector $\Sigma e_{i}=\lambda_{i} e_{i}$ and $\mathbb{E}(X)=\mu$:

$$ \begin{aligned} \text{Cov}(a^{T}X,e_{i}^{T}(X-\mu))&=a^{T}\lambda_{i} e_{i}-a^{T}\mu e_{i}^{T}\mu+a^{T} \mu e_{i}^{T}\mu \\ &=a^{T}\lambda_{i} e_{i}. \\ \end{aligned} $$


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