Subspaces

A subset $U$ of $V$ is called a subspace of $V$ if $U$ is also a vector space with the same additive identity, addition, and scalar multiplication as on $V$.

Conditions

How do we check whether the subset of a vector space is a subspace?

A subset $U$ of $V$ is a subspace of $V$ if and only if $U$ satisfies the following three conditions:

• Additive identity: $0\in U$.
• Closed under addition: $u,w\in U$ implies $u+w\in U$.
• Closed under scalar multiplication: $a\in \mathbb{F}$ and $u\in U$ implies $au\in U$.
  • More specifically, $a$ is a scalar from the field over which $V$ is defined.

How do we prove that a subset $U$ that satisfies those conditions is a vector space with the same additive identity, addition, and scalar multiplication as on $V$?

Proof

If $U$ is already known to be a subspace of $V$, then $U$ satisfies the three conditions above by the definition of vector space.

Conversely, suppose a subset $U$ satisfies the three conditions above. How do we know that $U$ is a vector space with the same properties as $V$ (hence being a subspace)? The three conditions tell us the following:

• The first condition ensures that the additive identity of $V$ is in $U$.
• The second condition ensures that addition makes sense on $U$.
• The third condition ensures that scalar multiplication makes sense on $U$.

If $u\in U$, then $-u$ (which is equal to $(-1)u$) is also in $U$ by the third condition above. Hence, every element of $U$ has an additive inverse in $U$.

The other parts of the definition of a vector space, such associativity and commutativity, are automatically satisfied for $U$ because they hold on the larger space $V$. Thus, $U$ is a vector space and hence is a subspace of $V$.

Q.E.D.

The set $\{0\}$ is the smallest subspace of $V$, and $V$ itself is the largest subspace of $V$. The empty set is not a subspace of $V$ because a subspace must be a vector space and hence must contain at least one element, namely, an additive identity.

The subspaces of ${\mathbb{R}}^2$ are precisely $\{0\}$, all lines in ${\mathbb{R}}^2$ containing the origin, and ${\mathbb{R}}^2$. The subspaces of ${\mathbb{R}}^3$ are precisely $\{0\}$, all lines in ${\mathbb{R}}^3$ containing the origin, all planes in ${\mathbb{R}}^3$ containing the origin, and ${\mathbb{R}}^3$.

Examples

Some of the examples below show the linear structure underlying parts of calculus. For example:

• Example 2 requires the result that the sum of two continuous functions is continuous.
• Example 4 requires the result that for a constant $c$, the derivative of $cf$ equals $c$ times the derivative of $f$

Subspace Example 1

Show that if $b\in \mathbb{F}$, then $\{(x_1,x_2,x_3,x_4)\in {\mathbb{F}}^4:x_3=5x_4+b\}$ is a subspace of ${\mathbb{F}}^4$ if and only if $b=0$.

We need to check whether if our set satisfies the three conditions of being a subspace: zero vector, closed under addition, closed under scalar multiplication.

Zero vector: The zero vector in ${\mathbb{F}}^4$ is $(0,0,0,0)$. For this vector to be in our set, it must satisfy $x_3=5x_4+b$. Thus, we have $0=5(0)+b$, which gives means for the zero vector to exist in our set, we must have $b=0$.

Closed under vector addition: Suppose we have $\mathbf{u}=(u_1,u_2,u_3,u_4)$ and $\mathbf{v}=(v_1,v_2,v_3,v_4)$. Then $u_3=5u_4+b$ and $v_3=5v_4+b$. The sum $\mathbf{u}+\mathbf{v}$ should be in our set. Thus, its third component of $\mathbf{u}+\mathbf{v}$ needs to satisfy:

$$u_3+v_3=5(u_4+v_4)+b$$

Substituting the expressions for $u_3$ and $v_3$, we get:

$$u_3+v_3=(5u_4+b)+(5v_4+b)$$

Thus, we need to have:

$$\begin{array}{rl}5(u_4+v_4)+b& =(5u_4+b)+(5v_4+b)\\ b& =2b\end{array}$$

which only holds if $b=0$.

Closed under scalar multiplication: Let $\mathbf{u}=(u_1,u_2,u_3,u_4)$ be in our set, such that $u_3=5u_4+b$. For any scalar $\lambda \in \mathbb{F}$, we have $\lambda \mathbf{u}=(\lambda u_1,\lambda u_2,\lambda u_3,\lambda u_4)$. For $\lambda \mathbf{u}$ to be in our set, the third component must then satisfy

$$\lambda u_3=5(\lambda u_4)+b$$

Substituting $u_3=5u_4+b$ into this, we then have

$$\begin{array}{rl}\lambda (5u_4+b)& =5(\lambda u_4)+b\\ 5\lambda u_4+\lambda b& =5\lambda u_4+b\end{array}$$

which holds if $\lambda b=b$. This is true for all $\lambda$ only if $b=0$.

Thus, our set is a subspace of ${\mathbb{F}}^4$ if and only if $b=0$. Otherwise, the set does not satisfy the conditions required to be subspace.

Subspace Example 2

Show that the set of continuous real-valued functions on the interval $[0,1]$ is a subspace of ${\mathbb{R}}^{[0,1]}$.

Recall that ${\mathbb{R}}^{[0,1]}$ is the set of all real functions on the interval $[0,1]$. This set includes every possible function that takes a real value for each $x\in [0,1]$. We can check whether the set of continuous real-valued functions meets the conditions of being a subspace of ${\mathbb{R}}^{[0,1]}$.

Zero vector: The zero vector in ${\mathbb{R}}^{[0,1]}$ is the function $f(x)=0$ for all $x\in [0,1]$. This is continuous on the interval $[0,1]$ because it is constant and equal to $0$ for all $x$. Therefore, the zero function is an element of the set of continuous real-valued functions on $[0,1]$.

Closed under addition: Let $f(x)$ and $g(x)$ be continuous real-valued functions on $[0,1]$, such that they are members of ${\mathbb{R}}^{[0,1]}$. The sum of two functions is continuous; since $f(x)$ and $g(x)$ are continuous on $[0,1]$, $h(x)=f(x)+g(x)$ is also continuous on $[0,1]$. Therefore, $h(x)$ also belongs to the set of continuous real-valued functions on $[0,1]$.

Closed under scalar multiplication: Let $f(x)$ be a continuous real-valued function $[0,1]$ and $a\in \mathbb{R}$ be a scalar. The scalar multiple of a continuous function is continuous; thus, $af(x)$ is also continuous on $[0,1]$. Therefore, $cf(x)$ belongs to the set of continuous real-valued functions on $[0,1]$.

This examples shows the linear structure underlying parts of calculus. It requires the result that the sum of two continuous functions is continuous.

Subspace Example 3

Show that the set of differentiable real-valued functions on $\mathbb{R}$ is a subspace of ${\mathbb{R}}^{\mathbb{R}}$.

The notation ${\mathbb{R}}^{\mathbb{R}}$ represents the set of all functions from the real numbers $\mathbb{R}$ to the real numbers $\mathbb{R}$. We want to show that the set of differentiable functions is a subspace of this set of all functions.

Zero element: The zero function $f(x)=0$ is differentiable on $\mathbb{R}$. Therefore, the set of differentiable real-valued functions on $\mathbb{R}$ fulfills the requirement of having the same zero element as ${\mathbb{R}}^{\mathbb{R}}$.

Closed under addition: Let $f(x)$ and $g(x)$ be differentiable real-valued functions on $\mathbb{R}$. We need to show that $h(x)=f(x)+g(x)$ is also differentiable on $\mathbb{R}$. We have

$$h^{\prime }(x)=\frac{d}{dx}[f(x)+g(x)]=f^{\prime }(x)+g^{\prime }(x)$$

Since $f^{\prime }(x)$ and $g^{\prime }(x)$ exist and are continuous, $h^{\prime }(x)$ also exists and is continuous, which means that $h(x)$ is differentiable. Thus, $h(x)$ also belongs to the set of differentiable real-valued functions on $\mathbb{R}$, satisfying the closure under addition requirement.

Closed under scalar multiplication: Let $f(x)$ be a differentiable real-valued function on $\mathbb{R}$. We need to show that $af(x)$ is also in the set of differentiable functions for some scalar $a\in \mathbb{R}$. We have

$$(a{f}^{\prime })(x)=\frac{d}{dx}(af(x))=a{f}^{\prime }(x)$$

Since $f^{\prime }(x)$ exists and is continuous, $a{f}^{\prime }(x)$ also exists and is continuous, which means that $af(x)$ is differentiable. Thus, $af(x)$ also belongs to the set of differentiable real-valued functions on $\mathbb{R}$, satisfying the closure under scalar multiplication requirement.

Subspace Example 4

Show that the set of differentiable real-valued functions $f$ on the interval $(0,3)$ such that $f^{\prime }(2)=b$ is a subspace of ${\mathbb{R}}^{(0,3)}$ if and only if $b=0$.

${\mathbb{R}}^{(0,3)}$ is the set of functions from $(0,3)$ to $\mathbb{R}$. To show that the set of differentiable real-valued functions $f$ on $(0,3)$ is a subspace of ${\mathbb{R}}^{(0,3)}$, we show that they share a zero element, and that the latter is closed under addition and scalar multiplication.

Zero element: The zero function in ${\mathbb{R}}^{(0,3)}$ is $f(x)=0$, which is differentiable on $(0,3)$. Thus, the set of differentiable real-valued functions on $(0,3)$ fulfills the requirement of having the same zero element as ${\mathbb{R}}^{(0,3)}$.

Closed under addition: Let $f(x)$ and $g(x)$ be differentiable real-valued functions on $(0,3)$ with $f^{\prime }(2)=g^{\prime }(2)=b$.

Let’s say that we have the sum $h(x)=f(x)+g(x)$. We have

$$h^{\prime }(x)=\frac{d}{dx}[f(x)+g(x)]=f^{\prime }(x)+g^{\prime }(x)$$

For the set of differentiable real-valued functions to be closed under addition, $h(x)$ must also be a member of this set, which means that we must have $h^{\prime }(2)=b$. Thus, we have:

$$\begin{array}{rl}{h}^{\prime }(2)& =f^{\prime }(2)+g^{\prime }(2)\\ b& =b+b\\ b& =2b\end{array}$$

which is only true if we have $b=0$.

Closed under scalar multiplication: Let $f(x)$ be a differentiable real-valued function on $(0,3)$ with $f^{\prime }(2)=b$. For the set of differentiable real-valued functions to be closed under scalar multiplication, $af(x)$ is also in the set of differentiable functions on $(0,3)$ for some scalar $a\in \mathbb{R}$. We have

$$(a{f}^{\prime })(x)=\frac{d}{dx}(af(x))=a{f}^{\prime }(x)$$

where $(a{f}^{\prime })(2)=b$. Thus, we would have:

$$\begin{array}{rl}(a{f}^{\prime })(2)& =a{f}^{\prime }(2)\\ b& =ab\end{array}$$

which is only true for all $a$ if $b=0$.

This examples shows the linear structure underlying parts of calculus. It requires the result that for a constant $c$, the derivative of $cf$ equals $c$ times the derivative of $f$.

Subspace Example 5

Show that the set of all sequences of complex numbers with limit 0 is a subspace of ${\mathbb{C}}^{\infty }$.

Let $S$ be the set of all sequences of complex numbers that converge to $0$. That is,

$$S=\{(a_n)|\underset{n\to \infty }{\lim }=0,a_n\in \mathbb{C}\text{ for all }n\}$$

We want to show that this is a subspace of ${\mathbb{C}}^{\infty }$, which is the space of all infinite sequences of complex numbers.

Zero element: The zero sequence $(0,0,0,\dots )$ is a sequence where every term is $0$. Since the sequence is constant and equal to $0$ for all $n$, it clearly converges to $0$. Therefore, the zero sequence is in $S$.

Closed under addition: Let $(a_n)$ and $(b_n)$ be two sequences in $S$. By definition, we have:

$$\underset{n\to \infty }{\lim }{a}_n=0,\underset{n\to \infty }{\lim }{b}_n=0$$

Consider the sequence formed by adding these two sequences, $(a_n+b_n)$. The limit of the sum of the sequences is:

$$\underset{n\to \infty }{\lim }(a_n+b_n)=\underset{n\to \infty }{\lim }{a}_n+\underset{n\to \infty }{\lim }{b}_n=0+0=0$$

Since ${\lim }_{n\to \infty }(a_n+b_n)=0$, the sequence $(a_n+b_n)$ is in $S$. Hence, $S$ is closed under addition.

Closed under scalar multiplication: Let $(a_n)$ be a sequence in $S$ and let $c\in \mathbb{C}$ be any complex number. We need to check if the sequence $(c\cdot a_n)$ is in $S$. The limit of the scalar multiple of the sequence is

$$\underset{n\to \infty }{\lim }(c\cdot a_n)=c\cdot \underset{n\to \infty }{\lim }{a}_n=c\cdot 0=0$$

Since ${\lim }_{n\to \infty }(c\cdot a_n)=0$, the sequence $(c\cdot a_n)$ is also in $S$. Therefore, $S$ is closed under scalar multiplication.