$\text{span}\{(1, 0), (0, 1)\} = \mathbb{R}^2$ because every $(a, b) = a(1, 0) + b(0, 1)$.

Also $\text{span}\{(1, 0), (0, 1), (1, 1)\} = \mathbb{R}^2$: the third vector is redundant since $(1, 1) = (1, 0) + (0, 1)$.

$\text{span}\{(2, 3)\} = \{t(2, 3) \mid t \in \mathbb{R}\} = \{(2t, 3t) \mid t \in \mathbb{R}\}$, which is the line through the origin with slope $3/2$.

$\text{span}\{(1, 0, 1), (0, 1, 1)\} = \{(a, b, a + b) \mid a, b \in \mathbb{R}\}$, which is the plane $z = x + y$ in $\mathbb{R}^3$.

$\text{span}\{(1, 2), (3, 6)\} = \text{span}\{(1, 2)\}$ because $(3, 6) = 3(1, 2)$. Two parallel vectors span only a line, not a plane. Geometrically, adding a redundant vector does not enlarge the span.

$\text{span}\{1, x, x^2\} = P_2(\mathbb{R})$, the space of polynomials of degree at most 2.

Alternatively, $\text{span}\{1, 1+x, 1+x+x^2\} = P_2(\mathbb{R})$ as well, since: $1 = 1$, $x = (1+x) - 1$, and $x^2 = (1+x+x^2) - (1+x)$.

The four matrices

$E_{11} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad E_{12} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad E_{21} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \quad E_{22} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$

span $M_{2 \times 2}(F)$, since any matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = aE_{11} + bE_{12} + cE_{21} + dE_{22}$.

Do the vectors $v_1 = (1, 2, 3)$, $v_2 = (4, 5, 6)$, $v_3 = (7, 8, 9)$ span $\mathbb{R}^3$?

We check: $v_3 = 2v_2 - v_1$ (since $(7,8,9) = 2(4,5,6) - (1,2,3)$). So $\text{span}\{v_1, v_2, v_3\} = \text{span}\{v_1, v_2\}$, which is a plane in $\mathbb{R}^3$. These three vectors do not span $\mathbb{R}^3$.

In $C(\mathbb{R})$, $\text{span}\{\sin x, \cos x\}$ is the set of all functions $a\sin x + b\cos x$ with $a, b \in \mathbb{R}$. This can also be written as $\{A\sin(x + \varphi) \mid A \geq 0, \varphi \in \mathbb{R}\}$.

$\text{span}\{1, x, x^2, x^3, \ldots\} = P(\mathbb{R})$, the space of all polynomials. This is an infinite-dimensional example: no finite subset spans all of $P(\mathbb{R})$.

In $C^\infty(\mathbb{R})$, $\text{span}\{e^x, e^{2x}, e^{3x}\}$ is a 3-dimensional subspace. These functions are linearly independent (as can be verified using a Wronskian determinant).

$\text{span}\{(1, 0, 0), (0, 1, 0)\}$ is the $xy$-plane in $\mathbb{R}^3$, not all of $\mathbb{R}^3$. The vector $(0, 0, 1)$ cannot be expressed as a linear combination of $(1, 0, 0)$ and $(0, 1, 0)$.

The subspace $W = \{(x, y, z) \in \mathbb{R}^3 \mid x + 2y - z = 0\}$ can be parametrized by setting $y = s$, $z = t$, giving $x = -2s + t$. So

$W = \{s(-2, 1, 0) + t(1, 0, 1) \mid s, t \in \mathbb{R}\} = \text{span}\{(-2, 1, 0), (1, 0, 1)\}.$