Using $\text{Cov}(A\mathbf{Y}+b) = A\text{Cov}(\mathbf{Y})A^{\prime}$ to prove that $\text{Cov}(\mathbf{Y})$ is nonnegative definite Suppose $\mathbf{Y}$ is a $n$-dimensional random vector, $A$ is a fixed $r \times n$ matrix, and $b$ is a fixed vector in $\mathbb{R}^n$. I have proven already that $$\mathbb{E}\left[A\mathbf{Y}+b\right] = A\mathbb{E}\left[\mathbf{Y}\right]+b\text{, }$$
$$\mathbb{E}\left[A\mathbf{Y}b\right] = A\mathbb{E}[\mathbf{Y}]b\text{,} $$
and
$$\text{Cov}\left(A\mathbf{Y}+b\right) = A\text{Cov}\left(\mathbf{Y}\right)A^{\prime}\text{,}$$
where $\text{Cov}(\mathbf{Y})$ denotes the (variance-)covariance matrix of $\mathbf{Y}$. 
The book says that using $\text{Cov}\left(A\mathbf{Y}+b\right) = A\text{Cov}\left(\mathbf{Y}\right)A^{\prime}$, we can show that $\text{Cov}\left(\mathbf{Y}\right)$ is nonnegative definite for any random vector $\mathbf{Y}$.
So we have for $v \in \mathbb{R}^n$,
$$v^{\prime} \text{Cov}\left(\mathbf{Y}\right) v = \text{Cov}\left(v^{\prime}\mathbf{Y}\right)\text{.}$$
Just looking at the matrix dimensions, I know that $v^{\prime}\mathbf{Y}$ is a $1 \times 1$ summation of terms. In particular,
$$v^{\prime}\mathbf{Y} = \begin{bmatrix}
v_1 & v_2 & \cdots & v_n 
\end{bmatrix}\begin{bmatrix}
y_{1} \\
y_2 \\
\vdots \\
y_n\end{bmatrix} = \sum\limits_{i=1}^{n}y_iv_i\text{.}$$
Just given my machinery above, I'm not sure how to find $\text{Cov}\left(\sum\limits_{i=1}^{n}y_iv_i\right)$ without it being very messy. 
Note that I already asked a very similar question here, but I would like to know what the author of the book is thinking by suggesting using $\text{Cov}\left(A\mathbf{Y}+b\right) = A\text{Cov}\left(\mathbf{Y}\right)A^{\prime}$.
 A: A square matrix $C = [C_{i,j}]$ is said to be positive 
semi-definite if and only if 
$$\sum_i \sum_j a_ia_j C_{i,j} \geq 0 ~ \forall a_i, a_j \in \mathbb R.
\tag{1}$$
It is called positive definite if the inequality in $(1)$ is a strict
inequality.
Now choose $A$ to be the row vector (a.k.a. $1\times n$ matrix)
$[a_1, a_2, \ldots, a_n]$ so that $A\mathbf Y$ is the univariate
random variable $Z= \sum_i a_iY_i$ and so 
$$\operatorname{cov}\left(A\mathbf Y+b\right)
= \operatorname{var}(Z) \geq 0.\tag{2}$$ But, we are given that
$$\operatorname{cov}\left(A\mathbf{Y}+b\right) = A\operatorname{cov}\left(\mathbf{Y}\right)A^{T}\tag{3}$$ where, for our
chosen $A$, the right side of $(3)$ is just
$\sum_i \sum_j a_i a_j \operatorname{cov}\left(\mathbf Y\right)_{i,j}$.
We conclude from $(2)$ and $(3)$ that
$$\sum_i \sum_j a_i a_j \operatorname{cov}\left(\mathbf Y\right)_{i,j}\geq 0~ \forall a_i, a_j \in \mathbb R,\tag{4}$$
that is, $\operatorname{cov}\left(\mathbf Y\right)$ is a positive
semi-definite matrix. 
Adapted from this previous answer of mine.
A: Recall any valid covariance matrix is nonneg def. So for any random vector $Y$ if $\mbox{Cov}(Y)$ is defined, it has $\mbox{Cov}(Y) \geq 0$. Furthermore, another property of NND matrices is that, for any real vector $A$, and any NND matrix $\Sigma$, $A^\prime \Sigma A$ is also NND.
A: In the result, plug $A = U'$, where $U$ is the matrix of eigenvectors. Writing the eigendecomposition $S = U \Lambda U^{T}$  reveals $Cov(UY)$ =$\Lambda$ which is a diagonal matrix
Now, the eigenvalues of a diagonal matrix are the diagonal entries. But, the diagonal entries are the variances and hence non-negative (this result can also be proved trivially) and therefore, the covariance matrix of $UY$ is positive definite.
Now use $Y_0 =UY$ and $A_{0} = U^{T}$ in the equation to get your result
