# Linear Algebra Done Right 2nd Edition – Chapter 1 Exercises 5(b), 5(c), 5(d), 6, 7

Chapter 1

5.(b)Determine if $U = {(x_1,x_2,x_3) \in \textbf{F}^3:x_1 + 2x_2 + 3x_3 = 4}$ is a subspace of $\textbf{F}^3$.

Solution: Since $0 + 0x_2 + 0x_3 \neq 4, (0,0,0) \notin U$. Thus $U$ is not a subspace of $\textbf{F}^3$.

5.(c)Determine if $U = {(x_1,x_2,x_3) \in \textbf{F}^3:x_1x_2x_3 = 0}$ is a subspace of $\textbf{F}^3$.

Solution: Since $x_1x_2x_3 = 0$ at least one of $x_i = 0$. Consider $(0,1,1)$ and $(1,0,1)$. Both are in $U$ but $(0,1,1) + (1,0,1) = (1,1,2) \notin U$. Thus, $U$ is not a subspace of $\textbf{F}^3$.

5.(d)Determine if $U = {(x_1,x_2,x_3) \in \textbf{F}^3:x_1 = 5x_3}$ is a subspace of $\textbf{F}^3$.

Solution:

Clearly $(0,0,0) \in U$.

Consider $(a,b,c)$ and $(x,y,z)$ both in $U$.
Then $(5c,b,c) + (5z,y,z) = (5c+5z, b+y, c+z) = ( 5(c+z), (b+y), (c+z) ) \in U$

Let $(x,y,z) = (5z, y, z) \in U$ and $a \in \textbf{F}$
Then $a(x,y,z) = (a5z, ay, az) = (5az, ay, az) \in U$. $U$ is a subspace of $\textbf{F}^3$

6. Give an example $U$ 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 1:

Let $U = \{ (x,y) \in \mathbb{R}^2 : x,y \text{ are rational } \}$ $U \neq \emptyset$ since $(0,0) \in U$.

If $(a,b), (c,d) \in U$ then $(a+c, b+d) \in U$ since the sum of rationals is rational. Thus, $U$ is closed under addition.

Also, $-a, -b$ are are rational since $a, b, -1$ are rational and the product of rationals is rational. Then $(-a,-b) \in U$. So $U$ is closed under taking additive inverses.

Now, let $a \in \mathbb{R}, a \text{ is irrational }$. Consider $(1,1) \in U$. Then $a(1,1) = (a,a) \notin U$ since the product of an irrational and a non-zero rational is an irrational. Thus, $U$ is not closed under scalar multiplication. $U$ is not a subspace of $\mathbb{R}^2$

Solution 2: This is the one in the solution manual.

Let $U = \{ (x,y) \in \mathbb{R}^2 : x,y \in \mathbb{Z} \}$ $U \neq \emptyset$ since $(0,0) \in U$.

Given $(a,b) \in U$ and $(c,d) \in U$ then $a+c \in \mathbb{Z}$ and $b+d \in \mathbb{Z}$ and $(a+c,b+d) \in U$.
Thus, $U$ is closed under addition.

Also, $-a, -b$ are integers since $a, b$ are integers. Then $(-a,-b) \in U$.
So $U$ is closed under taking additive inverses.

Consider $(1,1) \in U$ and $\frac{1}{2} \in \mathbb{R}$.
Then $\frac{1}{2} (1,1) = (\frac{1}{2}, \frac{1}{2}) \notin U$.
Thus, $U$ is not closed under scalar multiplication. $U$ is not a subspace of $\mathbb{R}^2$

7. 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:

Let $U = \{ (x,y) \in \mathbb{R}^2 : \text{ if } x = 0, y = 0 \text{ otherwise } y \neq 0 \}$

Just for fun, this is a different subset than the one found in the solution manual. $U \neq \emptyset$ since $(0,0) \in U$.

Let $(x,y) \in U$ and $a \in \mathbb{R}$. If $a = 0$ or $(x,y) = (0,0)$, then $a(x,y) = (0,0)$ otherwise $a(x,y) = (ax, ay) : ay \neq 0$. So $a(x,y) \in U$. Thus, $U$ is closed under scalar multiplication.

Consider $(10, 1)$ and $(-5, -1)$. Both are in $U$. But $(10, 1) + (-5, -1) = (5, 0) \notin U$. $U$ is not a subspace of $\mathbb{R}^2$.