Characterize $\mathcal{U}\subseteq\mathbb{R}^d$ such that $\mathbf{Var}(u^\top X) = 0$ for all $u\in\mathcal{U}$

Question

So I'm trying to solve the following question, my approach is at the end.

Let $X$ be a random vector with covariance matrix $\mathbf{Cov}(X) = \Sigma \in \mathbb{R}^{d \times d}$. Characterize the set $\mathcal{U} \subseteq \mathbb{R}^d$ that achieves

$$ \mathbf{Var}(u^\top X) = 0 \qquad\iff\qquad u \in \mathcal{U}.$$ Is $\mathcal{U}$ a subspace, if so what are its dimensions?

My approach - I'm not really sure how to characterize the set and this is my working to show that it is a subspace.

• Clearly $0 \in \mathcal{U}$.

• If $u_1 \in \mathcal{U}$ then $\alpha u_1 \in \mathcal{U}$ as well where $\alpha$ is a scalar.

• If $u_1, u_2 \in \mathcal{U}$ then

  \begin{align*} \mathbf{Var}[(u_1 + u_2)^\top X] &= (u_1+u_2)^\top \Sigma(u_1+u_2) \\ &= u_1^\top \Sigma u_1 + 2u_1^\top \Sigma u_2 + u_2^\top \Sigma u_2 \\ &= 2u_1^\top \Sigma u_2 \end{align*}

  This I cannot claim to be zero? So is this not a subspace?

Can someone please help me out with this and show me how to prove this is a subspace (if it actually is) and what it's dimension will be and lastly how to characterize it?

Answer 1

Let $\mathcal U$ be the set of vectors $u \in \Bbb R^d$ such that $Var(u^TX)=0$.

We have that $Var(u^TX)=u^T \Sigma u=0$. Since $X$ is a covariance matrix, it is a positive semidefinite matrix and we can write $0=u^T\Sigma u=u^T \Sigma^{\frac 12}\Sigma^{\frac 12}u=\lVert \Sigma^{\frac 12} u\rVert^2$, where $\Sigma^{\frac 12}$ is the square root of $\Sigma$, constructed by taking the square root of each eigenvalue in the spectral decomposition of $\Sigma$. By the properties of the norm, we have that $ \lVert \Sigma^{\frac 12} u\rVert^2 =0 \implies \Sigma^{\frac 12} u=0\implies \Sigma u=0$ so $u \in \ker(\Sigma)$, meaning that $\mathcal U \subset \ker(X)$.

You can easily see that $ \ker(\Sigma) \subset \mathcal U$ and thus, $\mathcal U=\ker(\Sigma)$.

Notice that when $\Sigma$ is positive definite, it is injective and $\ker(\Sigma)=\{ 0\}$ which matches the fact that $u^T \Sigma u>0$ for all non-zero vectors.

It then gives you the answer for the subspace question since $\ker(\Sigma)$ is a subspace of $\Bbb R^d$ and there are many ways to find its dimension, using rank–nullity theorem and the eigenvalues of $\Sigma$ for example.