math121b:01-31

Today, we first reviewed what our plan was. We try to answer a few questions they are

Our final goal for part I (material for midterm I) is to derive the Laplace operator in curvilinear coordinate, which you can find the formula here.

However, I am trying to present a more mathematical approach here, different from that of Boas. This involves introducing many concepts in linear algebra, useful in differential geometry (if you care about general relativity), and hopefully also in other parts of the engineer.

One possiblity is that, these math languages and viewpoints (ie. try not to use basis and components), will be quickly forgotten once you left this class, and never be encountered again in your career. I do think some exposure to these abstract way of thinking is helpful, hopefully one recognize that math is more than just a bunch of useful formula which you can use as a blackbox (hmm… math is also a verbose way to saying simple things), instead I personally think the main virtue of math is it allows one to see through common examples and see the common pattern and summarize that in a beautiful way.

Indeed, I understand, if one is just given some abstract definition, it would be too dry and too boring. So here is some example.

This example comes up when I am discussing with Henry from our class during office hour. I think it illustrate well the notion of vector space.

** The vector space of degree (at most) 3 polynomials **
Let $V$ denote the set of polynomials with degree at most 3
$$ V = \{ a x^3 + b x^2 + c x + d \mid a,b,c,d \in \R \}. $$
In particular, $a$ can be zero, so the degree of an element might be less than $3$.

So far, we have describe $V$ as a set. What is a set? What is a function? These are basic math terminologies, and are describe here. Basically, a set is a collection of elements. period. The points on the real line forms a set, denoted as $\R$; the atoms in my body is a set; the outcome of tossing a coin is a set $\{head, tail\}$. And a map between two sets, say from set $A$ to set $B$, to assign to each element in $A$, an element in $B$.

A vector space is a set with more features. It has 'addition of two vectors' and 'multiplication by a number', these two operation. So we see that our example $V$ indeed has those operations, and is a vector space.

Then, we give some example of map from $V$ to $\R$, i.e. an $\R$-valued function, for example,

- we can map everything in $V$ to $0$, this is an example of a constant function
- we can map everything in $V$ to $1$, another example of constant function.
- we can map an element $f(x)=a x^3 + b x^2 + c x + d$ to the real number $d$, denoted as $$ a x^3 + b x^2 + c x + d \mapsto d $$ where $\mapsto$ read 'maps to', is denoted the assignment of an element; whereas $\to$ as in $V \to \R$ is used to denote as map between two sets.
- we can map an element $f(x) \in V$, to $f(10)$. Indeed, $f(10)$ is a number, so this rule indeed defines a function from $V$ to $\R$.
- we can map $f(x)$ to $f(10)^3$.

Then, we checked, whether these functions are **linear** functions on $V$. The example 1,3,4, are linear, and the other two are not.

There are more examples. For example, for any $a \in \R$, we can define the 'evaluation map' $$ ev_a: V \to \R, \quad f(x) \mapsto f(a). $$ The example 3 above is $ev_0$ and example 4 is $ev_{10}$. And, we can take derivative first, then evaluate $$ f(x) \mapsto 3 f''(0). $$

Now, we take all such different linear function on $V$, and put them together into a set. That is the dual vector space of $V$.

$$V^* = \text{ the set of linear functions on $V$ } $$

Is this a crazy set? With all the various examples of linear functions? Well, not really. $V^*$ is a vector space at least. We also know its dimension, $4$, same as $\dim V$.

If you fix a basis of $V$ as follow $$ e_1 = 1, \quad e_2 = x, \quad e_3 = x^2, \quad e_4 = x^3 $$ then (I found a better way to explain, I think) the dual basis $\{h^i\}_{i=1}^4$ in $V^*$ is the following: say I want to describe $h^2$, then given an element $v \in V$, $h^2(v)$ is computed by first expand $v$ as linear combination of the $e_i$, then just pick out the coefficient in front of $e_2$. That is it, that is the value of $h^2(v)$, the coefficient in front of the $e_2$ in the expansion of $v$.

Hence $$h^2(a x^3 + b x^2 + c x + d) = h^2(d e_1 + c e_2 + b e_3 + a e_4) = c.$$

math121b/01-31.txt · Last modified: 2020/01/31 22:04 (external edit)