# Relationship between covariance matrix and its determinant

Let $X=(X_1, X_2,..., X_n)$ be random variables $$v_{ij} = cov(X_i, X_j) = E(X_i, X_j) - E(X_i)(X_j)$$

Show that the det of v is zero iff there are $a_1, a_2,..., a_n$ and b such that

$$P(a_1X_1 + a_2X_2 +... + a_nX_n = b) = 1$$

I'm not sure how to even begin this problem, so any leads would be great.

The determinant is zero if and only if the matrix is singular if and only if there exists $a$ such that $$Ca=0$$ where $C$ is the covariance matrix. The variance of $$y = a_1X_1 + a_2X_2 +... + a_nX_n$$ is $$a^t C a$$ which is zero.
So $y$ is almost surely constant and we are done.
Hint: Try computing $\text{Var}(a_1X_1+a_2X_2+\dots+a_nX_n)$. What does it mean if a random variable has zero variance?