LADR Exercises 1C

Problem 1

For each of the following subsets $S$ of ${\mathbb{F}}^3$, determine whether it is a subspace of ${\mathbb{F}}^3$.

• (a) $\{(x_1,x_2,x_3)\}\in {\mathbb{F}}^3:x_1+2x_2+3x_3=0$
• (b) $\{(x_1,x_2,x_3)\}\in {\mathbb{F}}^3:x_1+2x_2+3x_3=4$
• (c) $\{(x_1,x_2,x_3)\}\in {\mathbb{F}}^3:x_1{x}_2{x}_3=0$
• (d) $\{(x_1,x_2,x_3)\}\in {\mathbb{F}}^3:x_1=5x_3$

1(a) The zero vector in ${\mathbb{F}}^3$ is $(0,0,0)$. Since we have

$$0+2(0)+3(0)=0,$$

this zero vector also exists in the subset.

For addition, suppose we have $u,v\in S$ such that $u=(u_1,u_2,u_3)$ and $v=(v_1,v_2,v_3)$. Then, let’s say we have:

$$\begin{array}{rl}w& =u+v\\ (w_1,w_2,w_3)& =(u_1+v_1,u_2+v_2,u_3+v_3)\end{array}$$

For closure under addition, we need to have $w\in S$. So we check:

$$\begin{array}{rl}{w}_1+2w_2+3w_3& =(u_1+v_1)+2(u_2+v_2)+3(u_3+v_3)\\ & =(u_1+2u_2+3u_3)+(v_1+2v_2+3v_3)\\ & =0+0\\ & =0\end{array}$$

Thus, $w=u+v$ satisfies the condition for belonging in $S$, so we have closure under addition.

For scalar multiplication, let’s say we have $\lambda \in \mathbb{F}$ and $u\in S$ such that:

$$\lambda u=(\lambda u_1,\lambda u_2,\lambda u_3)$$

To check whether $\lambda u\in S$ for closure of scalar multiplication, we check

$$(\lambda u_1)+2(\lambda u_2)+3(\lambda u_3)=\lambda (u_1+2u_2+3u_3)=\lambda (0)=0$$

Thus, $\lambda u$ satisfies the condition for belonging in $S$, so we have closure under scalar multiplication as well.

1(b) This subset is not a subspace because the zero vector of ${\mathbb{F}}^4$ is not in it:

$$0+2(0)+3(0)\ne 4$$

1(c) The zero vector exists in the subspace, since $(0,0,0)$ has $0\cdot 0\cdot 0=0$. However, the additive identity is not valid since for some $u,v\in S$, we do not always have

$$(u_1+v_1)(u_2+v_2)(u_3+v_3)=0$$

For example, $(1,1,0)$ and $(0,1,1)$ both belong in the subset, but their sum $(1,2,1)$ does not.

1(d) The zero vector/additive identity for ${\mathbb{F}}^3$ exists in the subspace, since $(0,0,0)$ satisfies $x_1=5x_3$ with $0=5(0)$.

Closure under addition holds. For some $u,v\in S$ we have

$$\begin{array}{rl}w& =u+v\\ (w_1,w_2,w_3)& =(u_1+v_1,u_2+v_2,u_3+v_3)\end{array}$$

then, $w\in S$ if $w_1=5w_3$, which we can show with

$$\begin{array}{rl}{w}_1& =5w_3\\ u_1+v_1& =5(u_3+v_3)\\ u_1+v_1& =5u_3+5v_3\end{array}$$

For scalar multiplication, we have

$$\lambda u=(\lambda u_1,\lambda u_2,\lambda u_3)$$

which satisfies the closure property because

$$\begin{array}{r}\lambda u_1=5(\lambda u_3)\\ \lambda u_1=\lambda (5u_3)\end{array}$$

Problem 2

Verify all assertions about subspaces in Example 1.35.

This is done in Examples.

Problem 3

Show that the set of differentiable real-valued functions $f$ on the interval $(-4,4)$ such that $f^{\prime }(-1)=3f(2)$ is a subspace of ${\mathbb{R}}^{(-4,4)}$.

${\mathbb{R}}^{(-4,4)}$ is the set of functions from $(0,3)$ to $\mathbb{R}$. The zero function in ${\mathbb{R}}^{(-4,4)}$ is $f(x)=0$, which is differentiable on $(0,3)$. We also have:

$$\begin{array}{rl}{f}^{\prime }(-1)& =0\\ 3f(2)& =3(0)=0\end{array}$$

Thus, the zero function in ${\mathbb{R}}^{(-4,4)}$ is contained in $V$.

Next, for addition, if $f,g\in V$, then $f$ and $g$ are differentiable real-valued functions. So, $f+g$ must also be differentiable. Then:

$$(f+g{)}^{\prime }(-1)=f^{\prime }(-1)+g^{\prime }(-1)=3f(2)+3g(2)=3(f(2)+g(2))=3(f+g)(2)$$

from which we can conclude that $V$ is closed under addition.

For scalar multiplication, if $f\in V$ for any $\lambda \in \mathbb{R}$, then $f$ is differentiable real-valued functions. So, $\lambda f$ is differentiable too. Moreover,

$$(\lambda f{)}^{\prime }(-1)=\lambda f^{\prime }(-1)=\lambda (3f)(2)=3(\lambda f)(2)$$

This shows $V$ is closed under scalar multiplication.

Thus, we’ve shown that $V$ shares the same zero function/additive identity, is closed under addition, and is closed under scalar multiplication.

Problem 4

Suppose $b\in \mathbb{R}$. Show that the set of continuous real-valued functions $f$ on the interval $[0,1]$ such that ${\int }_0^{1}f=b$ is a subspace of ${\mathbb{R}}^{[0,1]}$ if and only if $b=0$.

Let us call the set $V$. If $V$ is a subspace of ${\mathbb{R}}^{[0,1]}$, then for any $f\in V$, we have ${\int }_0^{1}f=b$. For $V$ to be a subspace, any $kf$, where $k\in \mathbb{R}$, must also be in $V$. Hence

$$\begin{array}{rl}b& ={\int }_0^1(kf)\\ & =k{\int }_0^{1}f\\ & =kb\end{array}$$

which only happens if $b=0$.

If $b=0$, then for any $f,g\in V$ and $\lambda \in \mathbb{R}$, we have

$${\int }_0^1(f+g)={\int }_1^{0}f+{\int }_1^{0}g=0+0=0$$

which satisfies closure under addition, $f+g\in V$. Note that $f+g$ is a continuous real-valued function since $f$ and $g$ are.

Similarly,

$${\int }_0^1(\lambda f)=\lambda {\int }_0^{1}f=k0=0$$

and $\lambda f$ is a continuous real-valued function because $f$ is. Thus, we have closure under multiplication, $\lambda f\in V$.

Finally, we note the constant function (additive identity) $f\equiv 0$ in ${\mathbb{R}}^{[0,1]}$ is also the additive identity in $V$. Hence, $V$ is a subspace of ${\mathbb{R}}^{[0,1]}$.

Problem 5

Is ${\mathbb{R}}^2$ a subspace of ${\mathbb{C}}^2$?

${\mathbb{R}}^2$ is not a subspace of ${\mathbb{C}}^2$ because it does not satisfy closure under scalar multiplication. Specifically, we need to have $\lambda a\in {\mathbb{R}}^2$ for $\lambda \in \mathbb{C}$, $a\in {\mathbb{R}}^2$.

• Recall that for closure of multiplication where $U$ is a subspace of $V$, $\lambda$ is a scalar from the field over which $V$ is defined.

This is clearly not true:

$$\lambda =i,a=(1,1)⟶\lambda a=(i,i)\notin {\mathbb{R}}^2$$

Hence, ${\mathbb{R}}^2$ is not a subspace of the complex vector space ${\mathbb{C}}^2$.

Problem 6

• (a) Is $\{(a,b,c)\in {\mathbb{R}}^3:a^3=b^3\}$ a subspace of ${\mathbb{R}}^3$?
• (b) Is $\{(a,b,c)\in {\mathbb{C}}^3:a^3=b^3\}$ a subspace of ${\mathbb{C}}^3$?

6(a) The additive identity in ${\mathbb{R}}^3$ is $(0,0,0)$. This same additive identity exists in $V$.

For elements $x,y\in V$, we have:

$$\begin{array}{rl}x+y& =(a_x,b_x,c_x)+(a_y,b_y,c_y)\\ & =(a_x+a_y,b_x+b_y,c_x+c_y)\end{array}$$

For closure under addition, we need to have

$$(a_x+a_y{)}^3=(b_x+b_y{)}^3$$

or

$$a_x^3+3a_x^2{a}_y+3a_x{a}_y^2+a_y^3=b_x^3+3b_x^2{b}_y+3b_x{b}_y^2+b_y^3$$

This is true, since the $a_x^3=b_x^3$ and $a_y^3=b_y^3$ is only true if $a_x=b_x$ and $a_y=b_y$ (for real numbers). Thus, closure under addition is satisfied.

Closure under multiplication is easy to show:

$$\begin{array}{r}\lambda x=(\lambda a_x,\lambda b_x,\lambda c_x)\end{array}$$

and we have

$$\begin{array}{rl}(\lambda a_x{)}^3& =(\lambda b_x{)}^3\\ {\lambda }^3{a}_x^3& ={\lambda }^3{b}_x^3\\ a_x^3& =b_x^3\end{array}$$

Thus, we’ve shown that $V$ is indeed a subspace of ${\mathbb{R}}^3$.

6(b) We can show that the set is not a subspace of ${\mathbb{C}}^3$ by providing a counterexample. We have two elements in the subset, $x$ and $y$:

$$\begin{array}{rl}x& =\left(1,\frac{-1+\sqrt{3}i}{2},0\right)\in \{(a,b,c)\in {\mathbb{C}}^3:a^3=b^3\}\\ y& =\left(1,\frac{-1-\sqrt{3}i}{2},0\right)\in \{(a,b,c)\in {\mathbb{C}}^3:a^3=b^3\}\end{array}$$

Adding these two elements shows that the set is not closed under addition:

$$x+y=(2,-1,0)\notin \{(a,b,c)\in {\mathbb{C}}^3:a^3=b^3\}$$

Problem 7

Prove or give a counterexample: If $U$ is a nonempty subset of ${\mathbb{R}}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u\in U$ whenever $u\in U$), then $U$ is a subspace of ${\mathbb{R}}^2$.

We can say that