How can a random vector be nonsingular? I appreciate the help I have been getting on this site today. I had help on a proof that $\text{Cov}\left(\mathbf{Y}\right)$ is nonnegative definite for any random vector $\mathbf{Y}$. According to the textbook I have, it follows from this statement that $\mathbf{Y}$ is nonsingular if and only if $\text{Cov}\left(\mathbf{Y}\right)$ is positive definite.
Now this doesn't make sense to me, since $\mathbf{Y} \in \mathbb{R}^n$, and I thought that only square matrices were nonsingular? Also, I know that $\text{Cov}\left(\mathbf{Y}\right)$ is positive definite if and only if there exists a nonsingular (square matrix) $Q$ such that $$\text{Cov}\left(\mathbf{Y}\right) = QQ^{\prime}\text{.}$$ 
So are they perhaps actually saying that $\text{Cov}\left(\mathbf{Y}\right) = \mathbf{Y}\mathbf{Y}^{\prime}$, where $\mathbf{Y}$ is a nonsingular square matrix rather than vector?
 A: Your textbook may be referring to the distribution of $Y$ being a singular distribution: a distribution which doesn't have any point masses, but still its whole probability mass lies in a region of measure zero. This is kind of like $Y$ being in a lower-dimensional subset of $\mathbb R ^n$.
This is what I'd expect the statement to mean, but (as we'll see in a moment) this isn't an if and only if, just an only if. There may be some other conditions going on, or you may have misread it?
Let's call $\mathrm{Cov} Y$ just $\Sigma$ for brevity, and let $\mu = \mathbb E Y$.
First, if $\Sigma$ is singular, the distribution of $Y$ is singular.
As an example, suppose that $Y_1 \sim \mathcal{N}(0, 1)$ but $Y_2$ is always $0$. Then $\Sigma = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}$, a singular matrix, and $Y$ is concentrated on the $x$ axis of the plane (which is of measure zero).
In general, $\Sigma$ being singular means that there's some nonzero $q \in \mathbb R ^n$ with $\Sigma q = 0$. But note then that $\mathrm{Var}[ Y^T q] = q^T \Sigma q - (\mu^T q)^2 = -(\mu^T q)^2$, and since both variances and squared numbers must be nonnegative, we have both that $Y^T q$ has zero variance and $\mu^T q = 0$. Thus $Y$ lies only in the hyperplane $\{ y \mid q^T y = 0\}$, an $(n-1)$-dimensional subset of $\mathbb R ^n$.
This should make it clear that it's possible to have a singular distribution with a nonsingular covariance matrix, if $Y$ lies on any measure-zero subset other than a hyperplane through the origin.
For example, consider $Y$ to be uniformly distributed over the unit circle in $\mathbb{R}^2$; this has measure zero, but
$$\mathrm{Cov}(Y_1, Y_2) = \mathbb{E}[Y_1 Y_2] = \mathbb{E}_{Y_1}\left[ \mathbb{E}\left[ Y_2 \mid Y_1 \right] \right] = \mathbb{E}_{Y_1}\left[ 0 \right] = 0,$$
and clearly $\mathrm{Var}(Y_1) = \mathrm{Var}(Y_2) = v > 0$ for some $v$, so
$$\mathrm{Cov}[Y] = \begin{bmatrix}v & 0 \\ 0 & v\end{bmatrix}$$
which is nonsingular.
A: Matrices can be singular/non singular. Singular matrix basically means that the matrix is not full rank. For a $N \times P$ matrix with $N>P$, the matrix is non-singular if the rank of the matrix is $P$. By this notion I guess you can say vectors are never singular but that is a rather peculiar way of saying it. 
In terms of singular values, the result means you have $P$ non-zero singular values. Rank of ${\bf{YY}}'$ is equal to the number of non-zero singular values of ${\bf{Y}}$
When ${\bf{Y}}$ is a vector, ${\bf{YY}}'$ is always rank 1. If ${\bf{Y}}$ is a 2 $\times$ N on the other hand, based on the rank of ${\bf{Y}}$ you can tell if the covariance matrix is singular or not
