Why is $\operatorname{Cov}\left[v^TX,v_1^TX\right] = v^T\operatorname{Var}\left(X\right)v_1$?

Question

I came across the following result in one of the proofs exposed in my multivariate analysis course and I haven't been able to prove it :

Let $X=(X_1,\dots,X_p)$ be a random vector in $\mathbb{R}^p$. Then

\begin{equation} \label{1} \operatorname{Cov}\left(v^TX,v_1^TX\right) = v^T\Sigma v_1, \end{equation}

where

• $\Sigma := \operatorname{Var}(X) = \left[\operatorname{Cov}(X_i,X_j)\right]_{i,j=1,\dots,p}$
• $v \in \mathcal{S}^{p-1}$, where $\mathcal{S}^{p-1}$ is the unit sphere
• $v_1 := \underset{v \in \mathcal{S}^{p-1}}{\arg\max} \operatorname{Var}(v^TX)$

Note:

I'm not sure if the definitions of $v$ and $v_1$ should play somewhat of a role in the proof of the result (meaning that the statement might be proven for any vector $v$ and $v_1$) but it is how they are defined in the context of the course.

Answer 1

Using the hints provided to me in the comments, Here's the proof I came up with:

Let $\boldsymbol{v}=\begin{pmatrix} v_1, \dots, v_p \end{pmatrix}$ and $\boldsymbol{v_1}=\begin{pmatrix} v_{11}, \dots, v_{1p} \end{pmatrix}$. Using the covariance's bilinearity, we have

\begin{align} &\operatorname{Cov}\left(v^TX,v_1^TX\right) = \operatorname{Cov}\left(v_1X_1 +\dots+ v_pX_p, v_{11}X_1 +\dots+ v_{1p}X_p\right) \\\\ &= v_1\operatorname{Cov}\left(X_1,v_{11}X_1 +\dots+ v_{1p}X_p\right) + \dots + v_p\operatorname{Cov}\left(X_p,v_{11}X_1 +\dots+ v_{1p}X_p\right) \\\\ &= v_1\left[\operatorname{Cov}\left(X_1,X_1\right)v_{11} + \dots + \operatorname{Cov}\left(X_1,X_p\right)v_{11}\right] + \dots + v_p\left[\operatorname{Cov}\left(X_p,X_1\right)v_{11} + \dots + \operatorname{Cov}\left(X_p,X_p\right)v_{11}\right] \\\\ &= v_1\begin{bmatrix}\operatorname{Cov}\left(X_1,X_1\right) & \dots & \operatorname{Cov}\left(X_1,X_p\right)\end{bmatrix}\boldsymbol{v_1} + \dots + v_p\begin{bmatrix}\operatorname{Cov}\left(X_p,X_1\right) & \dots & \operatorname{Cov}\left(X_p,X_p\right)\end{bmatrix}\boldsymbol{v_1} \\\\ &= \boldsymbol{v}^T \begin{bmatrix}\operatorname{Cov}\left(X_1,X_1\right) & \dots & \operatorname{Cov}\left(X_1,X_p\right) \\ \vdots & \ddots & \vdots \\ \operatorname{Cov}\left(X_p,X_1\right) & \dots & \operatorname{Cov}\left(X_p,X_p\right) \end{bmatrix} \boldsymbol{v_1} \\\\ &= \boldsymbol{v}^T\Sigma \boldsymbol{v_1} \end{align}