linear algebra

Problem Show that for every $\alpha \in \mathbb{C}$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha + \beta = 0$ Solution Let $\alpha, \beta, \lambda \in \mathbb{F}$ and let $\alpha + \beta = 0$ and $\alpha + \lambda = 0$, then $$\beta = \beta + 0 = \beta + (\alpha + \lambda) \stackrel{P1A5}{=} (\beta + \alpha) + \lambda = 0 + \lambda = \lambda$$ Thus any additive inverse of $\alpha$ is unique, existence can be shown by seeing that if $\alpha = a + bi$, then $\beta = -a - bi$ is an additive inverse....

Problem Show that $(\alpha \beta) \lambda = \alpha (\beta \lambda)$ for all $\alpha, \beta, \lambda \in \mathbb{C}$ Solution Let $\alpha = a + bi$, $\beta = c + di$ and $\lambda = e + fi$. Then by the associativity of real numbers: $$(\alpha \beta) \lambda = \left((a + bi)(c + di)\right)(e + fi) \stackrel{D1.1}{=} \left(ac - df + (bc + ad)i\right)(e + fi)$$ $$\stackrel{D1.1}{=} (ace - bdf - bcf - adf) + (acf - bdf + bce + ade)i$$...

Problem Show that $(\alpha + \beta) + \lambda = \alpha + (\beta + \lambda)$ for all $\alpha, \beta, \lambda \in \mathbb{C}$ Solution Let $\alpha = a + bi$, $\beta = c + di$ and $\lambda = e + fi$. Then by the associativity of real numbers: $$(\alpha + \beta) + \lambda = \left((a + bi) + (c + di)\right) + (e + fi)$$ $$\stackrel{D1.1}{=} ((a + c) + (b + d)i) + (e + fi) \stackrel{D1....

Problem Show that $\alpha + \beta = \beta + \alpha$ for all $\alpha, \beta \in \mathbb{C}$. Solution Let $\alpha = a + bi$ and $\beta = c + di$ for real numbers $a,b,c$ and $d$. Then by commutativity of real numbers we must have: $$\alpha + \beta = (a + bi) + (c + di) \stackrel{D1.1}{=} (a + c) + (b + d)i$$ $$= (c + a) + (d + b)i \stackrel{D1....

Problem Find two distinct square roots of i. Solution First write i in complex exponential form, then take it’s square root and then use euler’s formula, as such: $$\sqrt{i} = \sqrt{e^{\frac{\pi}{2} i}} = \pm e^{\frac{\pi}{4}i} = \pm \cos\left(\frac{\pi}{4}\right) \pm i\sin\left(\frac{\pi}{4}\right) = \pm \frac{\sqrt{2}}{2}(1 + i)$$ Another approach is to let the square root of i equal a + bi and then solve for a and b as follows: $$ \sqrt{i} = a + bi \to i = (a + bi)^2 = a^2 - b^2 + 2abi$$ From which it must follow that: $$a^2 - b^2 = 0$$ $$2ab = 1$$ Therefore $a = b$ and $ab = 1/2$....