Problem Suppose $U_1$ and $U_2$ are subspaces of $V$. Prove that the intersection $U_1 \cap U_2$ is a subspace of $V$. Solution $U_1$ and $U_2$ are subspaces and therefore $0 \in U_1$ and $0 \in U_2$ implying that $0 \in U_1 \cap U_2$. If $x,y \in U_1 \cap U_2$, then $x + y \in U_1$ and $x + y \in U_2$ because these are subspaces. Therefore $x + y \in U_1 \cap U_2$ implying that $U_1 \cap U_2$ is closed under addition....

Problem A function $f: \mathbb{R} \to \mathbb{R}$ is called periodicif there exists a positive number $p$ such that $f(x) = f(x + p)$ for all $x \in \mathbb{R}$. Is the set of periodic functions from $\mathbb{R} \to \mathbb{R}$ a subspace of $\mathbb{R}^\mathbb{R}$? Explain. Solution No, the sum of two periodic functions is not necesarrily periodic. Proof: Take for example $f(x) = f(x+1)$ and $g(x) = g(x + \pi)$, which have periods of $1$ and $\pi$ respectively....

Problem Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $\mathbb{R}^2$. Solution With inspiration from problem 6 we construct the following subset of $\mathbb{R}^2$: $\lbrace (x,y) \in \mathbb{R}^2 | x^2 = y^2 \rbrace$. Then if $(x,y)$ is in the subset $\lambda (x,y) = (\lambda x, \lambda y)$ has the property that $(\lambda x)^2 = \lambda^2 (x^2) = \lambda^2 (y^2) = (\lambda y)^2$, that is - the subset is closed under scalar multiplication....

Problem Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \in U$ whenever $u \in U$), but $U$ is not a subspace of $\mathbb{R}^2$. Solution Consider $U^2 = \mathbb{Z}^2$, this is clearly a subset of $\mathbb{R}^2$ and is also closed under addition: $a, b \in \mathbb{Z}^2$ implies $a + b \in \mathbb{Z}^2$, also $a \in \mathbb{Z}^2 \to -a \in \mathbb{Z}^2$ since $\mathbb{Z}$ is closed under addition and negation....

Problem Solution Subset (a) In $\mathbb{R}$, the function $f(x) = x^3$ is injective and thus $a^3 = b^3 \iff a = b$. Therefore the subset of $\mathbb{R}^3$ we’re dealing with is $\lbrace {x,x,y} \in \mathbb{R}^3 | x,y \in \mathbb{R} \rbrace$ which is a subspace: $0 = (0,0,0)$ is in the subset, this is when $x=y=0$ $(x_1,x_1,y_1) + (x_2,x_2,y_2) \stackrel{D1.12}{=} (x_1 + x_2, x_1 + x_2, y_1 + y_2)$ and since we clearly have $x_1 + x_2 = x_1 + x_2$, the subset is closed under addition....