TITLE: Intuitive explanation of the Fundamental Theorem of Linear Algebra
QUESTION [17 upvotes]: 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.

REPLY [6 votes]: Suppose $V$ and $W$ are finite dimensional inner product spaces over $F$ (where $F$ is $\mathbb R$ or $\mathbb C$).  Let $T:V \to W$ be a linear transformation.
There are four subspaces that are naturally associated with $T$: $N(T), R(T), N(T^*)$, and $R(T^*)$.  (Here $T^*$ is the adjoint of $T$.  So $\langle T(x), y \rangle = \langle x, T^*(y) \rangle$ for all $x \in V, y \in W$.)
What Strang calls the fundamental theorem of Linear Algebra is the fact that
\begin{equation}
V = N(T) \perp R(T^*)
\end{equation}
and
\begin{equation}
W = N(T^*) \perp R(T).
\end{equation}
This theorem is easy to prove:
\begin{align}
& x \in N(T) \\
\iff & T(x) = 0 \\
\iff & \langle T(x),y \rangle = 0 \forall y \in W \\
\iff & \langle x, T^*(y) \rangle = 0 \forall y \in W \\
\iff & x \in R(T^*)^{\perp}.
\end{align}
This shows that $N(T)$ is the orthogonal complement of $R(T^*)$.
Similarly, $N(T^*)$ is the orthogonal complement of $R(T)$.
It's natural to seek bases for these four subspaces, and perhaps the "most natural" or nicest bases are the ones that arise in the SVD of $T$.

Edit:  It's interesting that, as @WillieWong pointed out, a similar theorem can be formulated in a more general setting, without using inner product spaces.
Let $V$ and $W$ be finite dimensional vector spaces over a field $F$ and let $T:V \to W$.  There are four subspaces naturally associated with $T$: $N(T), R(T), N(T^*)$, and $R(T^*)$.  (Now $T^*$ is the dual transformation $T^*:W^* \to V^*$.)
We may not have an inner product to work with, but we can still write
\begin{equation}
\langle T^*(w^*),v \rangle = \langle w^*,T(v) \rangle
\end{equation}
if we use the notation
\begin{equation}
\langle x^*,y \rangle := x^*(y)
\end{equation}
for $x^* \in V^*, y \in V$.
Moreover, $N(T)$ may not have an orthogonal complement, but it does have an annihilator, which is the analogous thing in a more general setting.  And we have that $R(T^*)$ is the annihilator of $N(T)$, and also that $N(T^*)$ is the annihilator of $R(T)$.  This is a more general form of Strang's "fundamental theorem".
Note that for any subspace $U$ of $V$, we have
\begin{equation}
\text{dim} \, U + \text{dim} \,U^{\perp} = \text{dim} \, V
\end{equation}
where $U^{\perp}$ is the annihilator of $U$.  In particular,
\begin{equation}
\text{dim} \, N(T) + \text{dim} \, R(T^*) = \text{dim} \, V.
\end{equation}
But we also know that
\begin{equation}
\text{dim} \, N(T) + \text{dim} \, R(T) = \text{dim} \, V.
\end{equation}
This shows that
\begin{equation}
\text{dim} \, R(T) = \text{dim} \, R(T^*).
\end{equation}
These easy but fundamental theorems and proofs appear in chapter 2 of Lax's book Linear Algebra and its Applications.  (See theorems 5, 5', and 6.)