nacho said:
Hi thanks for the responses everyone, I seem to have had a similar conceptual understanding, but would like to see it applied in a more general sense.
I have attached an example of what I mean.
How is it conclusive from the third and fourth line that the set W is closed under addition and multiplication?
I mean, all they've done is added them up and multiplied by scalar k. But can't you do that with any general set, and then turn around and say it is closed under scalar multiplication and addition?
Well, no, with some sets that doesn't work.
Here is a more sophisticated example:
We can form a vector space out of the set of real polynomial functions of degree less than or equal to two (this space is often called $P_2$ but it has dimension THREE), by defining, for:
$p_1(x) = a_0 + a_1x + a_2x^2$ and $p_2(x) = b_0 + b_1x + b_2x^2$ (here, $a_0a_1,a_2,b_0,b_1,b_2$ are all real numbers (constants)).
that:
$(p_1+p_2)(x) = (a_0 + b_0) + (a_1+b_1)x + (a_2+b_2)x^2$
$(cp_1)(x) = ca_0 + (ca_1)x + (ca_2)x^2$.
This has dimension 3, because one possible basis is $\{1,x,x^2\}$.
Now suppose we take the subset $S = \{p \in P_2: p(1) = 1\}$. We will see $S$ is not closed under addition. To do so, it suffices to find $f,g \in P_2$ with $f(1) = g(1) = 1$, but $(f+g)(1) \neq 1$.
Well one obvious such polynomial is the constant polynomial $f(x) = 1$, for all real $x$. Another is the linear polynomial $g(x) = x$.
Now $(f+g)(x) = 1 + x$, and we see that $(f+g)(1) = 1 + 1 = 2 \neq 1$.
So $S$ is not closed under addition, and as a result $S$ is not "big enough" to be a subspace.
HOWEVER, for ANY set $S$, we can always take $\text{span}(S)$ which includes ALL linear combinations (over our given field, usually the real numbers, but not always), and such a "span-set" will ALWAYS form a subspace.
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Let's look at a very simple vector space, the Euclidean plane $\Bbb R^2$ (if you understand this vector space well, you have a good platform to generalize from). Let's suppose we have a subset $S$ of the plane, that satisfies 3 properties:
1. Closure under addition:
if $(x,y),(x',y') \in S$ so is $(x+x',y+y')$
2. Closure under multiplication:
if $(x,y) \in S$ and $c$ is any real number, then $(cx,cy) \in S$.
3. $S$ has at least one vector (point) in it.
What would such a set look like?
Well two possibilities immediately present themselves:
a) The set consisting of just $\{(0,0)\}$. This is called a TRIVIAL subspace.
b) All of $\Bbb R^2$ itself (duh).
Let's suppose that we have two elements $(a,b),(a',b') \neq (0,0) \in S$ such that $(a',b') \neq (ca,cb)$ for any $c$.
I leave to you (or you can take it on faith)to show that in such a case $ab' - a'b \neq 0$ (hint: it's easier to show that if $ab' - a'b = 0$, that: $b' = cb, a' = ca$ for some $c$. Consider the cases $a = 0$ and $a \neq 0$ separately).
Now by (2), both $a'(a,b)$ and $(-a)(a',b')$ are in $S$, thus by (1), so is their sum:
$a'(a,b) + (-a)(a',b') = (aa',ba') - (aa',ab') = (0,a'b-ab')$.
Since $a'b - ab' \neq 0$ if we call this number $d$, we have:
$\frac{1}{d}(0,d) = (0,1) \in S$, by (2).
A similar argument (which you should provide) shows $(1,0) \in S$.
Finally, we see that any point $(x,y) = x(1,0) + y(0,1) \in S$, by taking (1) and (2) together.
So in this case, we see that two such points force $S$ to be all of $\Bbb R^2$, if $S$ is to satisfy our two closure conditions.
Are there any OTHER such sets $S$?
Let's try to find a minimal one "bigger than just the origin".
So we have at least one point (which is not the origin), say $(a,b) \in S$.
By (2) we have to have all points $(ca,cb) \in S$ for any real number $c$. Is this big enough? That is, we suppose:
$S = \{(ca,cb): c \in \Bbb R\}$, and since (2) is satisfied automatically, we just need to see if (1) is.
So, we pick two points in $S$, say:
$(x_1,y_1) = (c_1a,c_1b)$ and
$(x_2,y_2) = (c_2a,c_2b)$, and we see if $(x_1,y_1) + (x_2,y_2) \in S$.
Now $(x_1,y_1) + (x_2,y_2) = (c_1a,c_1b) + (c_2a,c_2b) = c_1(a,b) + c_2(a,b) = (c_1+c_2)(a,b) = ((c_1+c_2)a,(c_1+c_2)b)$.
Since $c_1+c_2$ is a real number if $c_1,c_2$ are, this shows the sum is indeed in $S$. So we do have closure (of addition).
What does such a set look like?
Since $(a,b) \neq (0,0)$ one of $a,b \neq 0$. Suppose that $a \neq 0$ (I leave it to you to puzzle out what happens if $a = 0$).
In this case, we have, for any point $(x,y) = (ca,cb) \in S$, that:
$y = \frac{b}{a}x$, that is, $S$ is the line through the origin with slope $\frac{b}{a}$.
What does this mean?
It means that any subset of the Euclidean plane closed under addition and scalar multiplication that has at least one point in it is either:
1. Just the origin
2. A line
3. The whole plane
This is why "linear algebra" is so-named. The "linearity" subspaces have is due to the "line-ness" (or in higher dimensions, the "plane-ness") of the subspaces.
