Vector Space Definition

For any field $F$ and positive integer $n$, the set $F^n = \{(a_1, \ldots, a_n) : a_i \in F\}$ is a vector space with:

Zero vector: $(0, 0, \ldots, 0)$

Additive inverse: $-v = (-a_1, \ldots, -a_n)$

The set $F[x]$ of all polynomials with coefficients in $F$ is a vector space with standard addition and scalar multiplication:

This is an infinite-dimensional vector space—no finite set of polynomials can span all of $F[x]$.

The set $F_n[x]$ of polynomials of degree at most $n$ is a finite-dimensional vector space of dimension $n + 1$.

Standard basis: $\{1, x, x^2, \ldots, x^n\}$.

The set of all $m \times n$ matrices with entries in $F$ is a vector space with entry-wise operations:

Dimension: $mn$ (the standard basis consists of matrices with exactly one entry equal to 1 and all others 0).

Let $\mathcal{F}(S, F)$ be the set of all functions from a set $S$ to a field $F$. With pointwise operations:

This is a vector space. If $S$ is infinite, the space is infinite-dimensional.

The set of all continuous functions $f: [a, b] \to \mathbb{R}$ is a vector space with pointwise operations. This is a subspace of the function space above.

Zero vector: The constant function $f(x) = 0$.

Key property: The sum of continuous functions is continuous, and a scalar multiple of a continuous function is continuous.

The set of all real sequences $(a_1, a_2, a_3, \ldots)$ forms a vector space with component-wise operations:

Special subspaces include:

• $\ell^2$: sequences with $\sum |a_n|^2 < \infty$
• $\ell^\infty$: bounded sequences
• $c_0$: sequences converging to 0

$\mathbb{C}$ is a 2-dimensional real vector space with basis $\{1, i\}$:

Note: $\mathbb{C}$ over $\mathbb{C}$ is 1-dimensional, but over $\mathbb{R}$ it's 2-dimensional. The field matters!

Let $V = \mathbb{R}^+$ (positive reals) with operations:

This is a vector space over $\mathbb{R}$!

• Zero vector: $1$ (since $x \cdot 1 = x$)
• Inverse of $x$: $1/x$ (since $x \cdot (1/x) = 1$)
• Isomorphism: $\log: (\mathbb{R}^+, \oplus, \odot) \to (\mathbb{R}, +, \cdot)$

Let $u = (1, 2, 3)$ and $v = (4, -1, 0)$ in $\mathbb{R}^3$. Then:

• $u + v = (1+4, 2+(-1), 3+0) = (5, 1, 3)$
• $3u = (3 \cdot 1, 3 \cdot 2, 3 \cdot 3) = (3, 6, 9)$
• $u - 2v = (1, 2, 3) + (-8, 2, 0) = (-7, 4, 3)$
• $-u = (-1, -2, -3)$

In $\mathbb{R}[x]$, let $p(x) = x^2 + 2x - 1$ and $q(x) = 3x - 5$. Then:

• $p + q = x^2 + 2x - 1 + 3x - 5 = x^2 + 5x - 6$
• $2p = 2x^2 + 4x - 2$
• $-p = -x^2 - 2x + 1$
• Zero vector: $0$ (the zero polynomial)

In $M_{2 \times 2}(\mathbb{R})$:

Zero vector: $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

In the space of continuous functions $C[0, 1]$:

Let $f(x) = x^2$ and $g(x) = \sin(\pi x)$. Then:

• $(f + g)(x) = x^2 + \sin(\pi x)$
• $(3f)(x) = 3x^2$
• $(-f)(x) = -x^2$
• Zero function: $h(x) = 0$ for all $x \in [0, 1]$

• ℤ over ℝ: Not a vector space because scalar multiplication by non-integers leaves ℤ. For example, $\frac{1}{2} \cdot 1 = \frac{1}{2} \notin \mathbb{Z}$.
• ℕ over ℝ: Not a vector space because natural numbers lack additive inverses.
• ℝ² with non-standard operations: Define $(a, b) \oplus (c, d) = (a + c, 0)$. This fails V1 (commutativity is okay) but fails V3: if $(a, b) \oplus (e, f) = (a, b)$, then $b = 0$ always, so no single zero works for all vectors.

• First quadrant: $\{(x, y) : x \geq 0, y \geq 0\}$ with standard operations. Fails: no additive inverse (e.g., $-(1, 1) = (-1, -1)$ is not in the set).
• ℚ over ℝ: Not a vector space. Scalar multiplication by irrationals leaves ℚ: $\sqrt{2} \cdot 1 = \sqrt{2} \notin \mathbb{Q}$.
• Polynomials of exact degree n: Not closed under addition! E.g., $(x^2 + x) + (-x^2 + 1) = x + 1$ has degree 1, not 2.
• Invertible matrices: Not a vector space! The zero matrix is not invertible, and the set is not closed under addition.

Let $S = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 = 1\}$ (the unit circle). Is this a vector space with standard operations?

Analysis:

• Zero? $(0, 0)$ has $0^2 + 0^2 = 0 \neq 1$. Not in $S$! ✗

Already fails. But also: $(1, 0) + (0, 1) = (1, 1)$ and $1^2 + 1^2 = 2 \neq 1$. Not closed under addition either.

Let $H = \{(x, y) \in \mathbb{R}^2 : y > 0\}$ (upper half-plane, excluding the x-axis).

Analysis:

• Zero? $(0, 0)$ has $y = 0$, not $> 0$. Not in $H$! ✗

Also: $(1, 1) + (-1, -2)$ would be needed for checking closure, but $(-1, -2) \notin H$. However, even for points in $H$: no additive inverses exist (if $y > 0$, then $-y < 0$).

Problem: Verify that $\mathbb{R}^2$ with standard operations is a real vector space.

Solution: We check all 8 axioms.

Problem: Is $S = \{(x, y) \in \mathbb{R}^2 : xy \geq 0\}$ a vector space?

Solution: No! Check closure under addition:

$(1, 1) \in S$ (since $1 \cdot 1 = 1 \geq 0$) and$(-1, 1) \in S$ (since $(-1) \cdot 1 = -1 < 0$)... wait, $(-1, 1) \notin S$.

Let's try: $(2, 1) \in S$ and $(-1, 2) \in S$.

Sum: $(2, 1) + (-1, 2) = (1, 3) \in S$. ✓

But: $(1, -1) \in S$ and $(-1, 1) \in S$.

Sum: $(1, -1) + (-1, 1) = (0, 0) \in S$. ✓

Actually, try $(1, 1)$ and $(-2, 1)$: both in $S$?

$1 \cdot 1 = 1 \geq 0$ ✓, $(-2) \cdot 1 = -2 < 0$ ✗

So $(-2, 1) \notin S$. The set $S$ consists of points in Q1, Q3, and axes.

Counter-example: $(1, 2) + (-3, 1) = (-2, 3)$. Check: $1 \cdot 2 = 2 \geq 0$ ✓, $(-3) \cdot 1 = -3 < 0$ ✗.

Need both in $S$: $(1, 2) \in S$, $(-3, -1) \in S$ (since $3 \geq 0$).

Sum: $(1, 2) + (-3, -1) = (-2, 1)$. Is $(-2)(1) = -2 \geq 0$? No!

Not closed under addition. Not a vector space.

Problem: Show that $W = \{(x, y, z) \in \mathbb{R}^3 : x + 2y - z = 0\}$ is a vector space.

Solution: $W$ is a subset of $\mathbb{R}^3$. Check subspace criteria:

1. Zero: $0 + 2(0) - 0 = 0$, so $(0, 0, 0) \in W$ ✓
2. Closed under +: If $x_1 + 2y_1 - z_1 = 0$ and $x_2 + 2y_2 - z_2 = 0$, then$(x_1 + x_2) + 2(y_1 + y_2) - (z_1 + z_2) = 0$ ✓
3. Closed under scalar mult: If $x + 2y - z = 0$, then$\alpha x + 2(\alpha y) - \alpha z = \alpha(x + 2y - z) = 0$ ✓

So $W$ is a subspace of $\mathbb{R}^3$, hence a vector space.

Problem: Show that $P_n(\mathbb{R})$ (polynomials of degree ≤ n) is a vector space.

Solution: $P_n(\mathbb{R}) \subseteq \mathbb{R}[x]$. Check subspace criteria:

1. Zero: The zero polynomial has degree $-\infty$ (or is considered to have any degree), so $0 \in P_n$ ✓
2. Closed under +: If $\deg(p) \leq n$ and $\deg(q) \leq n$, then $\deg(p + q) \leq \max(\deg p, \deg q) \leq n$ ✓
3. Closed under scalar mult: $\deg(\alpha p) = \deg(p) \leq n$ for $\alpha \neq 0$ ✓

Dimension: $n + 1$ (basis: $\{1, x, x^2, \ldots, x^n\}$).

Problem: Show that the set of symmetric $n \times n$ matrices is a vector space.

Solution: Let $S_n = \{A \in M_{n \times n}(\mathbb{R}) : A^T = A\}$.

1. Zero: $O^T = O$, so $O \in S_n$ ✓
2. Closed under +: If $A^T = A$ and $B^T = B$, then$(A + B)^T = A^T + B^T = A + B$ ✓
3. Closed under scalar mult: $(\alpha A)^T = \alpha A^T = \alpha A$ ✓

Dimension: $\frac{n(n+1)}{2}$ (diagonal + upper triangle entries).

Problem: Is $\{A \in M_{n \times n}(\mathbb{R}) : \text{tr}(A) = 0\}$ a vector space?

Solution: Yes! Check subspace criteria:

1. Zero: $\text{tr}(O) = 0$ ✓
2. Closed under +: $\text{tr}(A + B) = \text{tr}(A) + \text{tr}(B) = 0 + 0 = 0$ ✓
3. Closed under scalar mult: $\text{tr}(\alpha A) = \alpha \cdot \text{tr}(A) = \alpha \cdot 0 = 0$ ✓

Dimension: $n^2 - 1$.

Problem: Is the set of differentiable functions $f: \mathbb{R} \to \mathbb{R}$ a vector space?

Solution: Yes! With pointwise operations:

1. Zero: $f(x) = 0$ is differentiable ✓
2. Closed under +: Sum of differentiable functions is differentiable ✓
3. Closed under scalar mult: Scalar multiple of differentiable function is differentiable ✓

This is a subspace of all functions $\mathbb{R} \to \mathbb{R}$.

Problem: Show that solutions to $y'' + y = 0$ form a vector space.

Solution: Let $W$ be the set of solutions.

1. Zero: $y = 0$ satisfies $0'' + 0 = 0$ ✓
2. Closed under +: If $y_1'' + y_1 = 0$ and $y_2'' + y_2 = 0$, then$(y_1 + y_2)'' + (y_1 + y_2) = (y_1'' + y_1) + (y_2'' + y_2) = 0$ ✓
3. Closed under scalar mult: $(\alpha y)'' + \alpha y = \alpha(y'' + y) = 0$ ✓

Dimension: 2 (general solution: $y = c_1 \cos x + c_2 \sin x$).