Chapter 1

[bestsauce]

Chapter 1: Vector Spaces

1.A

• [x] Exercise 1
• [x] Exercise 2
• [x] Exercise 3
• [x] Exercise 4
• [x] Exercise 5
• [x] Exercise 6
• [x] Exercise 7
• [x] Exercise 8
• [x] Exercise 9
• [x] Exercise 10
• [x] Exercise 11
• [x] Exercise 12
• [x] Exercise 13
• [x] Exercise 14
• [x] Exercise 15
• [x] Exercise 16

1.B

• [x] Exercise 1
• [x] Exercise 2
• [x] Exercise 3
• [x] Exercise 4
• [x] Exercise 5
• [x] Exercise 6

1.C

• [x] Exercise 1
• [x] Exercise 2
• [x] Exercise 3
• [x] Exercise 4
• [x] Exercise 5
• [x] Exercise 6
• [x] Exercise 7
• [x] Exercise 8
• [x] Exercise 9
• [x] Exercise 10
• [x] Exercise 11
• [x] Exercise 12
• [ ] Exercise 13
• [x] Exercise 14
• [x] Exercise 15
• [x] Exercise 16
• [x] Exercise 17
• [x] Exercise 18
• [x] Exercise 19
• [x] Exercise 20
• [x] Exercise 21
• [x] Exercise 22
• [x] Exercise 23
• [x] Exercise 24

[bestsauce]

@MaxisJaisi Just wanted to see if you got some different examples for Ex7-8 and Ex19, here are mine:

• For Ex7: We can take as an example the lattice $\mathbf{Z}^2$.
• For Ex8: We can take the set $\big\{(x,y)\big| xy\geqslant0\big\}$.
• For Ex19: $U_1=\{0\}$ and $U_2=W$.

[MaxisJaisi]

Sure.

Same example for Ex 7. For Ex 8, take any two non-parallel lines passing through the origin. For Ex 19, I didn't quite get the simple counterexample you gave, so I again relied on the geometry of Euclidean spaces: let $V = \mathbf{R}^{3}$, and fix a plane passing through the origin, call it $W$. Then consider any other plane, $U_1$, passing through the origin which is distinct from the fixed plane. The sum of both planes would be all of $\mathbf{R}^{3}$. Again, take another plane distinct from $U_1$ and $W$, the sum would be $\mathbf{R}^{3}$, but $U_1 \neq U_2$. Or you could take a line and a plane (both passing through the origin), being careful that the line does not lie in the plane. The same story occurs, both spaces sum to $\mathbf{R}^{3}$.

[bestsauce]

@MaxisJaisi For Ex6 (b) I'm sure that there may be a simpler way, here's my solution:

Take $(\color{red}{1},\color{green}{1},p)\in\mathbf{C}^3$ and $\left(\color{red}{\dfrac{-1+\sqrt{3}i}{2}},\color{green}{1},c\right)\in\mathbf{C}^3$, they're all in that set, however their sum isn't since

$$\left(\color{red}{\dfrac{-1+\sqrt{3}i}{2}}+\color{red}{1}\right)^3=-1\neq 2^3=(\color{green}{1}+\color{green}{1})^3.$$

[MaxisJaisi]

@bestsauce Yes, the main idea is we can pick any $z \in \mathbf{C}$, and select any two distinct cube roots of $z$. Taking $z=1$ is probably the easiest way.