Why is this 'obviously' positive semi-definite? Here is a snapshot from a book I am studying. I learned all about positive semi-definiteness, and in fact I know that this matrix they are showing is in fact PSD. 

What I do not know is how they conclude it is PSD from the last part of the equation, when they factor the matrix into a vector times its transpose. 
How is it 'obvious' from here that this is PSD? 
Again, I know how to show the PSDness of this matrix, but I am not clear how/why they concluded the same thing, using the vector times vector^T in the end. 
Thanks.
 A: In general, let $a\in\mathbb R^n$ and $A=aa^T\in\mathbb R^{n\times n}$, as in the case of your matrix.
Clearly, $A$ is symmetric, as $A^T= (aa^T)^T=(a^T)^Ta^T=aa^T=A$.
$A$ is positive definite because of the fact that $(x,Ax)\ge 0$, for every $x\in\mathbb R^n$. 
Indeed
$$
(x,Ax)=x^TAx=x^Taa^Tx=(x^Ta)(a^Tx)=(x^Ta)^2\ge 0.
$$
Note that $x^Ta$ is nothing but the inner product of $x$ and $a$.
A: This follows from the observation that any matrix of the form $A = B^T B$ is automatically positive semi definite and is positive definite if $B$ has linearly independent columns(equivalently $B^T$ has linearly independent rows).
Proof: Let $x$ be any vector of the right dimension, then
$$
x^T A x= x^T B^T B x = y^T y \ge 0 ~~~\text{where $y=B x$}
$$
This shows that $A$ is semi definite. It will be definite if $y=0 \Rightarrow x=0$ which is equivalent to requiring $B$ to have linearly independent columns.
A: Another perspective on this: suppose $A$ has the form $A = uu^T$ for some vector $u$.  Then for any real vector $v$, we have
$$
Av = uu^Tv = u(u^Tv) = (u\cdot v)u 
$$
Where $u\cdot v$ represents a dot-product.
So, note that $Av$ will be $0$ if $v$ is perpendicular to $u$, and $Au = (u\cdot u)u$.  This determines all eigenvalues (and eigenvectors) of $A$.  Since all eigenvectors are non-negative, we conclude $A$ is PSD.
Moreover, it's probably good to know the following:

An $n \times n$ matrix $A$ is a rank-$k$ PSD matrix if and only if it can be expressed in the form $A = BB^T$ for some $n \times k$ matrix $B$.

A: I will just throw in another perspective. For a given matrix, if all the principle minors are non-negative, then the matrix is positive semi-definite. (Note: if you want positive definiteness, then you need to check only the leading principle minors.)
For the 2-by-2 matrix, this is easy to check. Here the principle minors are $x^2$, $y^2$, and $x^2 y^2 - x^2 y^2 = 0$. Clearly, $x^2$ and $y^2$ will always be positive, while the last one is zero. Hence the matrix is positive semi-definite.
