linear algebra

Problem Show that $(x + y) + z = x + (y + z)$ for all $x,y,z \in \mathbb{F}^n$ Solution Since addition in $\mathbb{F}^n$ is defined coordinate-wise, the result follows from the result shown in Problem 5 as such: $$(x + y) + z \stackrel{D1.12}{=} (x_1 + y_1, x_2 + y_2, \dots, x_n + y_n) + (z_1, z_2, \dots, z_n)$$ $$\stackrel{D1.12}{=} ((x_1 + y_1) + z_1, (x_2 + y_2) + z_2, \dots, (x_n + y_n) + z_n)$$ $$= (x_1 + (y_1 + z_1), x_2 + (y_2 + z_2), \dots, x_n + (y_n + z_n))$$ $$\stackrel{D1....

Problem Explain why there does not exist $\lambda \in \mathbb{C}$ such that $$\lambda (2 - 3i, 5 + 4i, -6 + 7i) = (12 - 5i, 7 + 22i, -32 - 9i)$$ Solution From the first coordinate and D1.17 we have: $$\lambda = \frac{2 - 3i}{12 - 5i} = \frac{12 + 5i}{12 + 5i}\frac{2 - 3i}{12 - 5i} = \frac{(24 + 15) + (10 - 36)i}{12^2 + 5^2} = \frac{3}{13} - \frac{2}{13}i$$ But from the second coordinate and D1....

Problem Find $x \in \mathbb{R}^4$ such that $$(4, -3, 1, 7) + 2x = (5, 9, -6, 8)$$ Solution First isolate $2x$ by subtracting the leftmost vector from both sides of the equation which can be done by applying D1.12, D1.16 and D1.17: $$2x = (5, 9, -6, 8) - (4, -3, 1, 7) = (1, 12, -7, 1)$$ Then divide by $2$ on both sides of the equation: $$x = \left(1/2, 6, -7/2, 1/2\right)$$ This is the solution....

Problem Show that $\lambda (\alpha + \beta) = \lambda \alpha + \lambda \beta$ for all $\alpha, \beta, \lambda \in \mathbb{C}$ Solution Let $\alpha = a + bi$, $\beta = c + di$ and $\lambda = e + fi$, then: $$\lambda (\alpha + \beta) = (e + fi)\left((a + bi) + (c + di)\right) \stackrel{D1.1}{=} (e + fi)\left((a + c) + (b + d)i\right)$$ $$\stackrel{D1.1}{=} (ea + ec - fb - fd) + (eb + ed + af + cf)i$$ $$= \left((ea - fb) + (ec - fd)\right) + \left((eb + af) + (ed + cf)\right)i$$ $$\stackrel{D1....

Problem Show that for every $\alpha \in \mathbb{C}$ with $\alpha \neq 0$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha \beta = 1$ Solution Let $\alpha,\beta,\lambda \in \mathbb{C}$ with $\alpha\beta = \alpha\lambda = 1$, then: $$\beta = \beta \cdot 1 = \beta(\alpha\lambda) \stackrel{P1A6}{=} (\beta\alpha)\lambda \stackrel{E1.4}{=} (\alpha\beta)\lambda = 1 \cdot \lambda = \lambda$$ Proving uniqueness of the multiplicative inverse $\beta$, existence was shown in P1A1.