Can someone explain intuitively what the Fundamental Theorem of Linear Algebra states? and why specifically it is called the above? Specifically, what makes it 'Fundamental' in the broad scope of the theory.
[Math] Intuitive explanation of the Fundamental Theorem of Linear Algebra
intuitionlinear algebra
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I'll try to answer the third question: the naming of the theorem in Wikipedia, as the References and External links sections suggest, totally follows from
Strang, Gilbert (1993), "The fundamental theorem of linear algebra", American Mathematical Monthly 100 (9): 848–855. doi:10.2307/2324660.
But you know, MAA is not a journal for frontier research. In particular, the above article is a (very informal) expository article; it doesn't even contain a clearly stated "theorem"—only some vague discussions are presented.
Now I'll state my personal opinion toward "the fundamental theorem." Fundamental theorems are surely important, but most often they are so easy to prove/so intuitive that after you've learned the subjects fairly well, they become your second nature—you never think about you are actually using some "theorem," and you never cite their names. (Did you ever cite The Fundamental Theorem of Calculus? Or Algebra? Unless you are trying to prove them or doing homework about their rather immediate implications, readers of your presentation might take it as an insult to their intelligence.) Therefore, since you're never going to cite their names, you don't need to discern which is which, unless you are a historian of mathematics.
The map $h : G/\ker f \to f(G)$ is essentially the same map as $f$, except that elements that were mapped to $0$ by $f$ are not taken into consideration anymore.
You can only do this when your subgroup is normal. As an example, consider the group $D_n = \langle r,s \, | \, r^n = s^2=1, srs = r^{-1} \rangle$. Take $n > 2$ for our example, and consider the two maps
$$ f_1 : D_n \to \mathbb Z/n \mathbb Z, \quad f_1(r^i s^j) = \overline i, \quad 0 \le i < n, \quad j \in \{0,1\} $$
$$ f_2 : D_n \to \mathbb Z/2 \mathbb Z, \quad f_2(r^i s^j) = \overline j, \quad 0 \le i < n, \quad j \in \{0,1\} $$ (The $\overline{i}$ means $i$ modulo $n$ ; I use the same notation for $j$ modulo $2$.) The first one is not even a group homomorphism ; we do not have the property that $f_1(s)f_1(r)f_1(s) = f_1(r^{-1})$ (check this ; this is where I use $n > 2$), which should have been true if $f_1$ was a group homomorphism. Intuitively, $f_1$ is an attempt to "mod out $s$". The reason why this attempt fails is explained by the fact that the subgroup $\langle s \rangle \subseteq D_n$ is not normal, i.e. $rsr^{-1} = r^2s \neq s$ when $n > 2$.
The second one, however, works just fine : it is a group homomorphism since $$ f_2(r^{i_1} s^{j_1} r^{i_2} s^{j_2}) = f_2(r^{i_1 - i_2} s^{j_1 + j_2}) = \overline{j_1 + j_2} = \overline{j_1} + \overline{j_2}. $$ This is illustrated by the fact that the subgroup $\langle r \rangle \subset D_n$ is normal, i.e. $r(r^i)r^{-1} = r^i$ and $s(r^i)s^{-1} = r^{-i}$.
So if we dismiss the information contained in a normal subgroup, we are fine ; this is because if $H \le G$ is a normal subgroup, then a product $gh$ where $g \in G$ and $h \in H$ will satisfy $gh = h'g$ where $h$ and $h'$ are elements of $H$. If you "don't read" $h$ and $h'$ in this equation, then you see $g$ on both sides. This is what being a normal subgroup means ; when we consider the elements of $H$ as behaving like the identity element in $G/H$, then we have a consistent group structure.
When you understand normal subgroups, you understand the projection map $\pi : G \to G/H$ for a normal subgroup $H$. The generalization $f : G \to K$ with $H = \ker f$ is not a big deal ; you can always replace $K$ by $f(G)$ (since $f(G)$ is a subgroup of $K$) and if you have read the proof of the first isomorphism theorem, you can always replace $f(G)$ by $G/H$, up to isomorphism. So your understanding of normal subgroups and the first isomorphism theorem reduces to understanding the projection map $\pi : G \to G/H$.
Hope that helps,
Best Answer
Imagine a projection, for example, from the whole $\mathbb{R}^3$ to the $x$-$y$ plane. It compresses each line that parallels to $z$-axis to a point on the plane. So there is a one-to-one relationship between the lines and the points. Notice that the lines are the translations of the $z$-axis -- that is also what quotient means. And the $z$-axis, is just the kernel of the projection, so we can see that $\operatorname{im} A\simeq V/\ker A$.
As for the dimension, the dimension of $\ker A$ measures how much we compress, while the dimension of $\operatorname{im} A$ measures how much we leave -- the amount we compress, plus the amount we leave, equals the whole thing, intuitively.