Consider $\mathbf{R}^2 = \{(x,y) | x,y \in \mathbf{R}\}$ with component wise addition and component wise scalar multiplication defined in the previous lecture. This is a vector space. Now, consider a line $L_m$ through the origin.

$$ \begin{align*} L_m &= \{(x,y) | y = mx\} = \{(x, mx) | x \in \mathbf{R}\}. \end{align*} $$

If we add two vectors in $L_m$, the result is in $L_m$.

$$ \begin{align*} (x, mx) + (\tilde{x}, m\tilde{x}) = (x + \tilde{x}, mx + m\tilde{x}) &= (x + \tilde{x}, m(x + \tilde{x})). \end{align*} $$

Similarly, scalar multiplication also preserves $L_m$. So $L_m$ seems to inherit the same structure of a vector space from $\mathbf{R}^2$. Let’s introduce the following definition

Definition

Let $W$ be a subset of a vector space $V$. $W$ is a subspace if

$\bar{0} \in W$.
For any $w_1, w_2 \in W$. $w_1 + w_2 \in W$. ($W$ is closed under addition)
For any $c \in \mathbf{R}$ and $w \in W$, $cw \in W$. ($W$ is closed under scalar multiplication).

Example 1: $[0, 1]$

Let $W = [0, 1] \subset \mathbf{R}$. It is not closed under addition. For example, for $w_1 = 1$ and $w_2 = 1$, $w_1 + w_2 \notin \mathbf{R}$. It is not closed under scalar multiplication either. $\bar{0} \in [0,1]$.

Example 2: $\mathbf{Z}$

$\bar{0} \in \mathbf{Z}$. It is closed under addition. But it is not closed under scalar multiplication because we defined scalar multiplication as $\{c(x, y) = (cx, cy) | c \in \mathbf{R}\}$. (Remember that the vector spaces we’re defining are over $\mathbf{R}$).

Subspaces are Vector Spaces

Theorem

If $W \subset V$ is a subspace and $V$ is a vector space, then $W$ is a vector space with the operations are inherited from $V$.

Proof: Let $V$ be a vector space and $W \subset V$. Since $V$ is a vector space, then we know it satisfies conditions (1)-(8) of vector spaces. We also know that $W$ satisfies properties (a)-(c) since it’s a subspace of $V$. To verify that $W$ is a vector space, we need to show that it satisfies conditions (1)-(8) of vector spaces.

Conditions (1)-(2) and (5)-(8) are inherited from $V$. For condition (3), we need a $\bar{0}$. But we already we know that there exists a zero element since it’s property (1) in the subspaces definition. For property (4), we need to prove that for each $w \in W$, there exists a $z \in W$ such that $w + z = \bar{0}$. But we know there is a $u \in V$ such that $w + u = \bar{0}$ since $V$ is a vector space. We also know that $u = (-1)w$ (In the previous lecture we proved that $u$ is unique). But $u=(-1)w$ must also be in $W$ since $W$ is closed under scalar multiplication. Therefore $W$ is a vector space as we wanted to show. $\blacksquare$

Example 3

Show that $W = \{(t_1 + t_2, t_1 - t_2, t_1) | t_1, t_2 \in \mathbf{R}\}$ is a vector space.

Proof: $W$ is a subset of $\mathbf{R}^3$ so by the previous theorem, it suffices to show that $W$ is a subspace of $\mathbf{R}^3$. For condition one, set $t_1$ and $t_2$ to 0 to get $\bar{0} = (0,0,0) \in W$. For any $w \in W$, we see that $w + \bar{0} = w + (0,0,0) = w$ as required. For condition 2, let $w_1, w_2 \in W$.

$$ \begin{align*} w_1 + w_2 &= (t_1 + t_2, t_1 - t_2, t_1) + (s_1 + s_2, s_1 - s_2, s_1) \\ &= (t_1 + t_2 + s_1 + s_2, t_1 - t_2 + s_1 + s_2, t_1 + s_1) \\ &= ((t_1 + s_1) + (t_2 + s_2), (t_1 + s_1) - (t_2 + s_2), (t_1 + s_1)). \end{align*} $$

Therefore, $w_1 + w_2 \in W$. For condition 3, let $w \in W$ and $c \in \mathbf{R}$,

$$ \begin{align*} cw &= c(t_1 + t_2, t_1 - t_2, t_1) \\ &= ( c(t_1 + t_2), ct_1 - ct_2, ct_1). \end{align*} $$

This is also clearly in $W$, Therefore, we can conclude that $W$ is a vector space. $\blacksquare$

Are Solution Sets Vector Spaces?

From the previous example, $W$ has the form of a solution set of a linear system so does this mean that every solution set is a vector space? The answer is no. Consider $W' = \{t_1 + t_2, t_1 - t_2, t_1 + 1\}$. $W'$ is not a subspace of $\mathbf{R}^3$. We don’t have a zero vector $\bar{0}$ in $W$. To show that, we need to prove that there isn’t a solution for the system $\{t_1 + t_2, t_1 - t_2, t_1 + 1\}$ where $t_1+t_2 = 0$, $t_1 - t_2 = 0$ and $t_1 + 1 = 0$.

Example 4: Polynomials of Degree at Most $k$

Consider $P_k$ which is the set of all polynomials of degree at most $k$. $P_k$ is a subspace of $P_{k+1}$. To show this, we know that $P_{k+1}$ is a vector space so now we just need to prove the three conditions of subspaces. The zero polynomial is in $P_k$. For condition 2, if we take any two polynomials in $k$, then

$$ \begin{align*} (a_kx^k + a_{k-1}x^{k-1} + ... + a_0) + (b_kx^k + b_{k-1}x^{k-1} + ... + b_0) = \\ (a_k + b_k)x^k + (a_{k-1}+b_{k-1})x^{k-1} + ... + (a_0+b_0). \end{align*} $$

The result is in $P_k$ which means that $P_k$ is closed under addition. For condition 3, take a polynomial in $P_k$ and a scalar $c \in \mathbf{R}$, then

$$ \begin{align*} c(a_kx^k + a_{k-1}x^{k-1} + ... + a_0) = (ca_k)x^k + (ca_{k-1})x^{k-1} + ... + (ca_0). \end{align*} $$

This is also in $P_k$ which means that $P_k$ is closed under scalar multiplication. Therefore, $P_k$ is a subspace of $P_{k+1}$.

Example 4: Continuous Functions

The claim is that $C^0(\mathbf{R})$ is a subspace of $F(\mathbf{R})$. For the zero vector, $\bar{0}(x) = 0$ is a continuous function so it belongs to $C^0(\mathbf{R})$. For condition 2, if $f$ and $g$ are continuous then we know that $f+g$ is also continuous (from calculus, needs to be proved). For condition 3, a scalar multiplied by $f$ is also continuous.

Example 5: Continuous Functions

Consider the map transpose: $M_{m \times n} \rightarrow M_{n \times m}$.

$$ \begin{align*} A = (a_{ij}) \rightarrow A^t = (a_{ij} = a_{ji}). \end{align*} $$

Example:

$$ \begin{align*} \begin{bmatrix} a & b \\ c & d \end{bmatrix}^t = \begin{bmatrix} a & c \\ b & d \end{bmatrix} \end{align*} $$

Definition

$A \in M_{n \times n}$ is symmetric if $A = A^t$.

We can show that the set of symmetric matrices in $M_{n \times n}$ is a subspace. Let the set of symmetric matrices in $M_{n \times n}$ be $S$. To show that $S$ is a subspace, we’ll verify the three conditions

$\bar{0} = \begin{bmatrix} 0... &... 0 \\ 0... &... 0 \end{bmatrix}^t$ is symmetric and therefore it is in $S$
The sum of two symmetric matrices is symmetric and therefore it is in $S$ and so $S$ is closed under addition. The sum is symmetric because given two symmetric matrices $A$ and $B$, their component wise sum is $a_{ij}+b_{ij}$. But then we know that
$$ \begin{align*} a_{ij} + b_{ij} = a_{ji} + b_{ji}. \end{align*} $$
Since both matrices are symmetric. But this implies that $A+B = (A+B)^t$.
For any $c \in \mathbf{R}$ and $s_1 \in S$, $cS$ is a symmetric matrix and therefore it is in $S$. So ($S$ is closed under scalar multiplication).

References:

Video Lectures from Math416 by Ely Kerman.

Lecture 5: Subspaces

Example 1: \([0, 1]\)

Example 2: \(\mathbf{Z}\)