Does $x^{T}y$ = $\|x\|\|y\|$? Does $x^{T}y$ = $\|x\|\|y\|$? I know that $x^{T}x=\|x\|^{2}$, so I assumed that $x^{T}y$ = $\|x\|\|y\|$. However, in Schwarz inequality, $|x^{T}y| \leq \|x\|\|y\|$ so I got confused there. Also, I'm not quite sure what the single bars around $|x^{T}y|$ mean. 
 A: Take $x=(1,0)$ and $y=(0,1)$.
Then
$$
\|x\|=\|y\|=1 \quad\text{and}\quad x^Ty=(1,0)\cdot(0,1)=0.
$$
Thus in this case $x^Ty\ne\|x\|\,\|y\|$.
A: $x^Ty \ne \Vert x \Vert \Vert y \Vert$ in general.  A family of counterexamples to $x^Ty = \Vert x \Vert \Vert y \Vert$ is provided by taking $x = (a, b)^T$ and $y = (-bc, ac)^T$ for $a, b, c \ne 0$.  Then $\Vert x \Vert = \sqrt{a^2 + b^2}$, $\Vert y \Vert = \vert c \vert \sqrt{a^2 + b^2}$, whence $\Vert x \Vert \Vert y \Vert = \vert c \vert (a^2 + b^2)$ but $x^Ty = -abc + abc = 0$.  The single bars denote the absolute value function; since $x^Ty$ is a scalar, this makes sense.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: This is not true in general. In Euclidean space we have a nice description of $x^Ty$, though: it's the dot product $x \cdot y$, and from your linear algebra (or perhaps vector calculus) class you have $$x \cdot y = \lVert x\rVert \lVert y\rVert \cos(\theta)$$ where $\cos(\theta)$ is the angle between the two vectors. Thus in this case, $x^Ty = \lVert x\rVert\lVert y\rVert$ if and only if $x$ and $y$ are collinear! This is also true for a general inner product space (see the statement of the Cauchy-Schwarz inequality, paticularly the bit on when equality holds.)
