what are the differences between alignment and colinear? Im reading a chapter talking about orthogonal complement of dual space in optimization by vector space. And the author introduced a definition of Alignment as following: 

where X* means the dual of X and < x , x* > denote a functional as following:

What are the differences between the alignment and colinear? Can you give me more examples? Thank you^_^
P. S. 


*

*Riesz-Frechet:





*Riesz Representation Theorem:

 A: The dual space is not an element of the underlying space, so, in general, it doesn't make sense to talk about collinear. Informally, however, alignment is basically the same idea, taking the different spaces into account.
For Hilbert spaces, the Riesz representation theorem shows that the space and its dual are isomorphic (antilinear, depending on your taste), so we can identify the dual and the underlying space as in the case of $\mathbb{C}^n$, for example.
For an example, take $X= l_1$, then $X^* = l_\infty$ (or, at least, identified with). Define $\sigma \in X^*$ by $\sigma(x) = \sum_n x_n$. Then $s$ is aligned with $x$ iff $s(x) = \|s\| \|x||$. Since $\|s\| = 1$, $s$ and $x$ are aligned iff $s(x) = \|x\|$, which is easily seen to be equivalent to $x_n = |x_n|$ for all $n$. Note that (as an element of $l_\infty$) we have $s=(1,1,...)$, so we cannot identify $s$ with any element of $X$.
Luenberger's book has an example involving the dual of $C[0,1]$ which illustrates the point a little more.
As an analog of the above, consider $f$ a continuous linear functional on $C[0,1]$ given by $f(x) = \int_0^1 x(t)dt$ (that is, the average value of $x$).
It is easy to see that $\|f\| = 1$, so to look for aligned points, we solve $f(x) = \|x\|$. Since $x$ is continuous, we see that $\int_0^1 x(t) dt = \max_t |x(t)|$ iff $x(t) = \max_t |x(t)|$ for all $t$, that is the constant functions whose value is non-negative.
Another example (well, not exactly) in $C[0,1]$ is $f(x) = \int_0^\frac{1}{2} x(t) dt- \int_\frac{1}{2}^1 x(t) dt$. A little work shows that $\|f\| = 1$ and $|f(x)| < \|x\|$ for all $x$, and so there are no non-zero vectors that are aligned with $f$.
