Intuitive explanation of the Fundamental Theorem of Linear Algebra

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.

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Read en.wikipedia.org/wiki/Fundamental_theorem_of_linear_algebra and the article by Strang in the Monthly cited in that page. – lhf Feb 9 '11 at 9:40
Argh, I rather disliked this "fundamental theorem". The whole statement in terms of orthogonality of the kernel and the co-image is obscuring the real statement about vector spaces and their duals. The fundamentality of the theorem (the "non-obvious" relation between kernel, cokernel, image, and coimage) is only due to the setting of working on $\mathbb{R}^n$ and identifying with its dual space using the standard inner product. (The other half of the theorem [rank-nullity], however, is a very fundamental one, I think. It makes the definition of dimension sensible.) – Willie Wong Feb 9 '11 at 18:03
It all boils down to this: The only invariant of a finitely generated vector space is its dimension. – Christian Blatter May 17 '11 at 17:28

Imagine a projection, for example, from the whole $\mathbb{R}^3$ to the $x$-$y$ plane. It compress 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.