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 ${\mathbb{R}}^2$.

Let’s say we have $U=\{(x,y)\in {\mathbb{R}}^2:x,y\in \mathbb{Z}\}$. $U$ is not empty.

If we have $(x_1,y_1)\in U$ and $(x_2,y_2)\in U$, then $x_1,x_2,y_1,y_2\in \mathbb{Z}$. The sums of each pair are integers, $x_1+y_1\in \mathbb{Z}$ and $x_2+y_2\in \mathbb{Z}$, so we have

$$(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)\in U$$

which means that $U$ is closed under addition.

Similarly, since $(-x_1,-y_1)\in U$, $U$ is closed under additive inverses.

However, $U$ is not closed under scalar multiplication if we have fractional scalars. For example, while $(1,1)\in U$, we do not have $\frac{1}{2}(1,1)\in U$.

Therefore, $U$ is not a subspace of ${\mathbb{R}}^2$.

Problem 8

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 ${\mathbb{R}}^2$.

An example nonempty subset is

$$U=\{(x,y)\in {\mathbb{R}}^2:x=0\text{ or }y=0\}$$

This is closed under scalar multiplication because for some elements $u,v$ and a scalar $\lambda$, we have

$$\begin{array}{rl}u=(x,0)& ⟶\lambda u=(\lambda x,0)\\ v=(0,y)& ⟶\lambda u=(0,\lambda y)\end{array}$$

where we have some $x,y\in \mathbb{R}$.

However, $U$ is not closed under addition because

$$u+v=(x,0)+(y,0)=(x,y)\notin U$$

Hence, $U$ is not a subspace of ${\mathbb{R}}^2$.

Problem 9

A function $f:\mathbb{R}\to \mathbb{R}$ is called periodic if 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.

Consider two periodic functions from $\mathbb{R}$ to $\mathbb{R}$:

$$\begin{array}{rl}f(x)& =\sin \sqrt{2}x\\ g(x)& =\cos x\end{array}$$

Then, for this to be a subspace, it needs to be closed under addition, such that the sum of the above functions

$$h(x)=\sin \sqrt{2}x+\cos x$$

is also periodic.

For $h(x)$ to be periodic, we need to have some $p$ such that $h(x)=h(x+p)$ for all $x\in \mathbb{R}$. Then, we must have

$$h(0)=1=h(p)=h(-p)$$

This means we must have:

$$1=\sin \sqrt{2}p+\cos p=-\sin \sqrt{2}p+\cos p$$

which implies that $\sin \sqrt{2}p=0$ and $\cos p=1$. We know from $\cos p=1$ that we must have $p=2\pi k$ where $k\in \mathbb{Z}$. However, $\sin \sqrt{2}p=0$ implies that $\sqrt{2}p=\ell \pi$, where $\ell \in \mathbb{Z}$. Hence,

$$\sqrt{2}=\frac{\ell \pi }{2k\pi }=\frac{1}{2k}\in \mathbb{Q}$$

which is not possible since $\sqrt{2}\notin \mathbb{Q}$.

Thus, this cannot be a subspace since our contradiction above shows it is not closed under addition.

Problem 10

Suppose $V_1$ and $V_2$ are subspace of $V$. Prove that the intersection $V_1\cap V_2$ is a subspace of $V$.

Additive identity: Since $V_1$ and $V_2$ are subspaces of $V$, they both contain the zero vector. Then, the zero vector must also be in the intersection $V_1\cap V_2$.

Closure under addition: Let $u,v\in V_1\cap V_2$. Then, this means that $u\in V_1,u\in V_2,w\in V_1,w\in V_2$. Since $V_1$ is a subspace, it is closed under addition, so $u+v\in V_1$. Since $V_2$ is also a subspace, $u+v\in V_2$. Since $u+v$ is in both $V_1$ and $V_2$, we have

$$u+v\in V_1\cap V_2$$

which shows that $V_1\cap V_2$ is closed under addition.

Closure under scalar multiplication: Let $u\in V_1\cap V_2$ and $\lambda \in \mathbb{F}$. We know that $u\in V_1$ and $u\in V_2$. Since $V_1$ and $V_2$ are both subspaces, they are closed under scalar multiplication, such that

$$\lambda u\in V_1,\lambda u\in V_2$$

Thus, we also have $\lambda u\in V_1\cap V_2$. Therefore, $V_1\cap V_2$ is closed under scalar multiplication.

Problem 11

Prove that the intersection of every collection of subspaces of $V$ is a subspace of $V$.

This seems to be a trivial extension of Problem 10 above. The idea is just that if a given element exists in the intersection of a collection, it must also exist in each of the subspaces in the collection. Conversely, a sum of two such elements must also exist in each of the subspaces for each of them to be closed under addition; since they exist in every subspace in the collection, they’re in the intersection. Same goes for scalar multiplication, and all of the subspaces must share the same additive identity since they are all subspaces of $V$.

Problem 12

Prove that the union of two subspaces of $V$ is only a subspace if and only if one of the subspace is contained in the other.

Suppose $U$ and $W$ are two subspace of $V$ and $U\cup V$ is a subspace of $V$.

Let’s say that $U$ is not a subset of $W$ and $W$ is not a subset of $U$. Consider elements $u\in U\setminus W$ and $w\in W\setminus U$. For $U\cup W$ to be closed under addition, their sum $u+w$ must belong to $U\cup W$. Hence, we must have $u+w\in U$ or $W$.

If $u+w\in U$, then $w=(u+w)-u\in U$, leading to a contradiction because we defined $w\in W\setminus U$.

If $u+w\in W$, then $u=(u+w)-w\in W$, leading to a contradiction because we defined $u\in U\setminus W$.

Therefore, if $U\cup W$ is a subspace of $V$, we must have $U\subset W$ or $W\subset U$.

Problem 13

Prove that the union of three subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces contains the other two.

Problem 14

Suppose

$$\begin{array}{rl}U& =\{(x,-x,2x)\in {\mathbb{F}}^3:x\in \mathbb{F}\}\\ W& =\{(x,x,2x)\in {\mathbb{F}}^3:x\in \mathbb{F}\}\end{array}$$

Describe $U+W$ using symbols, also give a description of $U+W$ that uses no symbols.

Symbolic: To find $U+W$, we consider all possible sums of elements from $U$ and $W$. Let $u\in U$ and $w\in W$, so:

$$\begin{array}{r}u=(x,-x,2x),x\in \mathbb{F}\\ w=(y,y,2y),y\in \mathbb{F}\end{array}$$

Adding $u$ and $w$:

$$u+w=(x+y,-x+y,2(x+y))$$

Let $s=x+y$ and $t=y-x$. Then we have:

$$x=\frac{s-t}{2},y=\frac{s+t}{2}$$

Therefore, every vector in $U+W$ can be expressed as:

$$u+w=(s,t,2s):s,t\in \mathbb{F}$$

In set form:

$$U+W=\{(x,y,2x)\in {\mathbb{F}}^3|x,y\in \mathbb{F}\}$$

Without symbols: $U+W$ is the set of all vectors in ${\mathbb{F}}^3$ whose third element is twice the first coordinate. This means that $U+W$ is a plane in ${\mathbb{F}}^3$ defined by $z=2x$.

Problem 15

Suppose $U$ is a subspace of $V$. What is $U+U$?

$U+U$ is the set of all possible sums of two elements from $U$. Since $U$ is a subspace of another space, it is closed under addition, the sum of two elements from $U$ should still be in $U$. Thus, $U+U=U$.

Problem 16

Is the operation of addition on the subspaces of $V$ commutative? In other words, if $U$ and $W$ are subspace of $V$, is $U+W=W+U$?

The addition on the subspaces is commutative because the addition of individual elements is commutative as well. For $x\in U$ and $y\in W$, we have

$$x+y\in U+W=y+x\in W+U$$

because addition in $V$ is commutative. This implies that $U+W\subset W+U$, and $W+U\subset U+W$. Hence, $U+W=W+U$.

Problem 17

Is the operation of addition on the subspaces of $V$ associative? In other words, if $V_1,V_2,V_3$ are subspaces of $V_s$ is

$$(V_1+V_2)+V_3=V_1+(V_2+V_3)$$

This is similar to the question above. If $x_1\in V_1,x_2\in V_2,x_3\in V_3$, we have:

$$(x_1+x_2)+x_3=x_1+(x_2+x_3)$$

since $x_1,x_2,x_3\in V$, which is a vector space (must then satisfy associative addition).

Problem 18

Does the operation of addition on the subspaces of $V$ have an additive identity? Which subspaces have additive inverses?

If some subspace $U$ is an additive identity, for for any subspaces $W$ of $V$, we need to have $U+W=W$. This means $U\subset W$; the only possibility is $U=0$.

Suppose some subspace of $W$ of $V$ has additive inverses, then there exists some $S$ of $V$ such that $W+S=0$. This can only happen when $W=0$ since $W\subset W+S$.

Problem 19

Prove or give a counterexample: If $V_1,V_2,U$ are subspaces of $V$ such that

$$V_1+U=V_2+U$$

then $V_1=V_2$.

A counterexample: Let $V=V_1=\{(a,b)\in {\mathbb{R}}^2:a,b\in \mathbb{R}\}$, $V_2=\{(a,0)\in {\mathbb{R}}^2:a\in \mathbb{R}\}$ and $U=\{(0,b)\in {\mathbb{R}}^2:b\in \mathbb{R}\}$. Then:

$$V_1+U=V_2+U$$

but $V_1\ne V_2$.

Problem 20

Suppose

$$U=\{(x,x,y,y)\in {\mathbb{F}}^4:x,y\in \mathbb{F}\}$$

Find a subspace of $W$ of ${\mathbb{F}}^4$ such that ${\mathbb{F}}^4=U\oplus W$.

A possible solution is:

$$W=\{(0,p,0,q)\in {\mathbb{F}}^4:p,q\in \mathbb{F}\}$$

This can be checked by considering the condition for direct sums of 2 subspaces, which is that if $U+W$ is a direct sum, we must have $U\cap W=\{0\}$. This is indeed true, as there are no elements in $W$ other than $0$ that have the same 1st and 2nd coordinate, and the same 3rd and 4th coordinate.

Problem 21

Suppose

$$U=\{x,y,x+y,x-y,2x\in {\mathbb{F}}^5:x,y\in \mathbb{F}\}$$

Find a subspace $W$ of ${\mathbb{F}}^5$ such that ${\mathbb{F}}^5=U\oplus W$.

$U$ only allows us to choose two coordinates arbitrarily; the rest of the coordinates depend on the first two. Thus, we take $W$ in which the first two coordinates are zero and the last three are variables:

$$W=\{0,0,p,q,r\in {\mathbb{F}}^5:p,q,r\in \mathbb{F}\}$$

We can show that $U\cap W=\{0\}$ by considering that, for some $\lambda =(a,b,c,d,e)\in U\cap W$, it must have $a=b=0$ because this is true for all elements in $W$. Following this fact, for $\lambda$ to be in $U$, it must have

$$\lambda =(x,y,x+y,x-y,2x)=(0,0,0+0,0-0,2(0))=(0,0,0,0,0)$$

Hence, we must have $\lambda =0$ and $U\cap W=0$.

Problem 22

Suppose

$$U=\{(x,y,x+y,x-y,2x)\in {\mathbb{F}}^5:x,y\in \mathbb{F}\}$$

Find three subspace $W_1,W_2,W_3$ of ${\mathbb{F}}^5$, none of which equal $\{0\}$, such that ${\mathbb{F}}^5=U\oplus W_1\oplus W_2\oplus W_3$.

We can have:

$$\begin{array}{r}{W}_1=\{(0,0,p,0,0)\in {\mathbb{F}}^5:p\in \mathbb{F}\}\\ W_2=\{(0,0,0,q,0)\in {\mathbb{F}}^5:q\in \mathbb{F}\}\\ W_3=\{(0,0,0,0,r)\in {\mathbb{F}}^5:r\in \mathbb{F}\}\end{array}$$

Problem 23

Prove or give a counterexample: If $V_1,V_2,U$ are subspaces of $V$ such that

$$V=V_1\oplus U\text{and}V=V_2\oplus U$$

then $V_1=V_2$.

A counterexample: Let $V={\mathbb{R}}^2$, $V_1=\{(x,0)\in {\mathbb{R}}^2:x\in \mathbb{R}\}$, $V_2=\{(0,y)\in {\mathbb{R}}^2:y\in \mathbb{R}\}$, and $U=\{(z,z)\in {\mathbb{R}}^2:z\in \mathbb{R}\}$. Then, we have

$$V=V_1\oplus U\text{and}V=V_2\oplus U$$

however $V_1\ne V_2$.

Problem 24

A function $f:\mathbb{R}\to \mathbb{R}$ is called even if

$$f(-x)=f(x)$$

for all $x\in \mathbb{R}$. A function $f:\mathbb{R}\to \mathbb{R}$ is called odd if

$$f(-x)=-f(x)$$

for all $x\in \mathbb{R}$. Let $V_e$ denote the set of real-valued even functions on $\mathbb{R}$ and let $V_o$ denote the set of real-valued odd functions on $\mathbb{R}$. Show that ${\mathbb{R}}^{\mathbb{R}}=V_e\oplus V_o$.

Given any $f\in {\mathbb{R}}^{\mathbb{R}}$ (the set of functions from $\mathbb{R}$ to $\mathbb{R}$), define an even part and an odd part

$$f_e(x)=\frac{f(x)+f(-x)}{2},f_o(x)=\frac{f(x)-f(-x)}{2}$$

for all $x\in \mathbb{R}$. Then, $f_e,f_o\in {\mathbb{R}}^{\mathbb{R}}$.

For all $x\in \mathbb{R}$, we have

$$f_e(-x)=\frac{f(-x)+f(x)}{2}=\frac{f(x)+f(-x)}{2}=f_e(x)$$

and

$$f_o(-x)=\frac{f(-x)-f(x)}{2}=-\frac{f(x)-f(-x)}{2}=-f_o(x)$$

hence $f_e\in V_e$ and $f_o\in V_o$.

We also have

$$f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}=f_e(x)+f_o(x)$$

so $f=f_e+f_o$. By definition, $f_e+f_o\in V_e+V_o$, so $f\in V_e+V_o$. Since we can choose $f$ arbitrarily from ${\mathbb{R}}^{\mathbb{R}}$, and we have shown that $f\in V_e+V_o$, it follows that every function in ${\mathbb{R}}^{\mathbb{R}}$ is in $V_e+V_o$, so we have ${\mathbb{R}}^{\mathbb{R}}=V_e+V_o$.

Then, to show ${\mathbb{R}}^{\mathbb{R}}=V_e\oplus V_o$, we need to show that $V_e\cap V_o=\{0\}$. Let $f\in V_e\cap V_o$. Then, we must have $f(x)=f(-x)$ since $f\in V_e$, and $f(x)=-f(-x)$ since $f\in V_o$ for all $x\in \mathbb{R}$. If we sum up $f(x)=f(-x)$ and $f(x)=-f(-x)$, we get $f(x)=0$ for all $x\in \mathbb{R}$. Hence, $f=0$, which implies $V_e\cap V_o=\{0\}$.