Prove some properties about the vectors i, i+j in R2

Let $S = \{ \mathbf{i}, \mathbf{i} + \mathbf{j} \}$ , where $\mathbf{i} = (1,0)$ and $\mathbf{j} = (0,1)$ are the unit coordinate vectors in $\mathbb{R}^2$ .

Prove that $S$ is linearly independent.
Prove that $\mathbf{j}$ is in the span of $S$ .
Express $3 \mathbf{i} - 4 \mathbf{j}$ as a linear combination of the vectors in $S$ .
Prove that $L(S) = \mathbb{R}^2$ .

Proof. We have
$x \mathbf{i} + y (\mathbf{i} + \mathbf{j}) = 0 \quad \implies \quad x+y = 0 \quad \text{and} \quad y= 0.$

But then, $y =0$ implies $x = 0$ . Hence, $x = y= 0. \qquad \blacksquare$
Proof. Let $x = -1$ and $y = 1$ . Then,
$x \mathbf{i} + y (\mathbf{i} + \mathbf{j}) = (-1,0) + (1,1) = (0,1) = \mathbf{j}. \qquad \blacksquare$
We have $3 \mathbf{i} - 4 \mathbf{j} = (3,-4)$ . Then,
$(3,-4) = x \mathbf{i} + y (\mathbf{i} + \mathbf{j}) \quad \implies \quad x+y = 3, \quad y = -4.$

Thus, $x = 7$ and $y = -4$ is a solution.
Proof. Since $\mathbf{i}$ and $\mathbf{i} + \mathbf{j}$ are two linearly independent vectors in $\mathbb{R}^2$ we have by Theorem 12.10 (page 466 of Apostol) that $L(S) = \mathbb{R}^2. \qquad \blacksquare$