If you have an orthonormal basis $\{ e_{k} \}_{k=1}^{N}$ on a finite-dimensional space, such as what you would obtain with Gram-Schmidt, then every vector $x$ is expressed as
$$
x = \sum_{k} (x,e_{k})e_{k}.
$$
This extends to $L^{2}[0,2\pi]$ using $e_{k} =\frac{1}{\sqrt{2\pi}}e^{ikx}$:
$$
f = \sum_{k}(f,e_{k})e_{k}
$$
The importance of this basis is that it consists of eigenvectors of $Lf=\frac{1}{i}\frac{d}{dx}$ because $Le_{k}=ke_{k}$. So this basis diagonalizes the differentiation operator. Finally, the same thing holds in a continuous sense on $L^{2}(\mathbb{R})$ with
\begin{align}
f & = \int_{k} (f,e_{k})e_{k} \\
& = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(t)e^{-ikt}dt\right)e^{ikx}dk
\end{align}
This is a generalization rather than a precise extension because $e_{k}=\frac{1}{\sqrt{2\pi}}e^{ikx}$ is not--strictly speaking--an eigenfunction of $L=\frac{1}{i}\frac{d}{dx}$ because $e_{k} \notin L^{2}(\mathbb{R})$ due to the fact that the function is not square integrable. However, for every $\delta > 0$, the following is square integrable
$$
e_{k,\delta}=\frac{1}{\sqrt{2\pi}}\int_{k-\delta}^{k+\delta}e_{k}(x)dk,
$$
and, in the norm of $L^{2}$, it becomes closer and closer to an eigenvector with eigenvalue $k$ as $\delta\downarrow 0$:
$$
\|Le_{k,\delta}-ke_{k,\delta}\| < \delta\|e_{k,\delta}\|.
$$
So the Fourier transform is the coefficient function and the expansion of $f$ looks very much like a "continuous" (i.e., integral) expansion of $f$ in approximate eigenfunctions of the differentiation operator. (The $e_{k,\delta}$ are even mutually orthogonal if the intervals $(k-\delta,k+\delta)$ do not overlap.)
As a final note, to make this generalization more precise,
$$
\|x\|^{2} = \sum_{k}|(x,e_k)|^{2}
$$
also holds for the continuous orthogonal expansion:
$$
\|f\|^{2} = \int_{k} |(x,e_{k})|^{2}dk.
$$
This is how Parseval saw it, who is the person after whom Parseval's identity is named:
$$
\int_{-\infty}^{\infty}|f(x)|^{2}dx = \int_{-\infty}^{\infty}|\hat{f}(k)|^{2}dk.
$$
It's not at all obvious that your two vectors span all of the plane, so this is a great question.
A usual proof goes like this:
"The span of those vectors is a 2-dimensional subspace of a 2-dimensional space. Hence, by a decomposition theorem usually proved pretty early, it must equal the whole space."
That's not very helpful, even if you know the decomposition theorem I'm talking about. So here's an alternative approach.
If your two vectors both have $0$ in the first entry, then they are in fact not independent. So at least ONE of them has a nonzero first entry. Call that $a$, and the other one $b$. In your example,
$$
a = \pmatrix{1\\9}, b = \pmatrix{11\\25}
$$
Now form a linear combination of $a$ and $b$ that looks like this:
$$
a' = \frac{1}{a_1} a, b' = b. \tag{1}
$$
Now $a'$ has a $1$ as its first entry. Observe that any combination of $a'$ and $b'$ is ALSO a combination of $a$ and $b$ (just substitute in formula 1).
In your example, $a' = a, b' = b$, because $a_1$ just happens to be $1$.
Now let
$$
b'' = b' - b'_1 a' = \pmatrix{11\\25} - 11 \pmatrix{1\\9} = \pmatrix{0\\-74}
$$
Now let's get that bottom entry of $b''$ nicer: replace
$$
a''' = a''; b''' = \frac{1}{b''_2}b'' = \frac{1}{-74} \pmatrix{0\\-74}= \pmatrix{0\\1}
$$
And finally, let $$
a'''' = a''' - a'''_2 b''' = \pmatrix{1\\9} - 9 \pmatrix{0\\1} = \pmatrix{1\\0}.
$$
And now we have shown that the two standard basis vectors can be written as linear combinations of $a$ and $b$, and therefore (by back-substituting), any combination of the standard basis vectors can also be written as a combination of $a$ and $b$. Hence all of $\Bbb R^2$ is in the span of $a$ and $b$.
Best Answer
If you are wondering why the linear system $$ 2x-y=1,\\x+y=5\tag{1} $$ and the so called column form (by Strang) $$ x\begin{bmatrix} 2\\1 \end{bmatrix} +y\begin{bmatrix} -1\\1 \end{bmatrix}= \begin{bmatrix} 1\\5 \end{bmatrix}\tag{2} $$
above are the same, then the short answer would be
Here is Strang's own explanation in his textbook: Consider the example: