# Vector Spaces

Mathematics I · Linear Algebra · https://tryals.app/learn/mathematics-i/vector-spaces

## Abstracting Vectors

Unit 3 treated vectors as arrows and tuples. **Linear algebra** abstracts away the picture and keeps only the operations. A **vector space** is any set $V$ of objects equipped with two operations, **addition** and **scalar multiplication**, satisfying a list of **axioms**. Anything obeying the axioms _is_ a vector space.

The axioms formalize familiar rules:

| Property | Rule | Description |
| --- | --- | --- |
| **Addition** | Commutative & associative | Order and grouping do not matter |
| **Zero vector** | $\mathbf{0} + v = v$ | Leaves vectors unchanged |
| **Inverse** | $v + (-v) = \mathbf{0}$ | Yields the zero vector |
| **Scaling** | $c(u+v) = cu + cv$ | Distributes and scales properly |

**Common pitfall:** Thinking a vector must be an **arrow** or a **tuple of numbers**. In linear algebra, a vector is _any_ element of a set satisfying the axioms. What matters is that you can **add** and **scale** them consistently.

## Examples and Power

The power of abstraction is that many _non-arrow_ objects satisfy the axioms and are true **vector spaces**:

| Space Type | Elements | Addition & Scaling |
| --- | --- | --- |
| $\mathbb{R}^n$ | Familiar tuples | Coordinate-wise |
| **Polynomials** | Degree $\le n$ | Combined by like powers |
| **Matrices** | Fixed size $m \times n$ | Added and scaled entry-wise |
| **Functions** | $f : \mathbb{R} \to \mathbb{R}$ | Added and scaled pointwise |
| **Solutions** | Differential equations | Homogeneous linear solutions |

Because all these share the same axioms, _every_ theorem proved for abstract vector spaces applies to _all_ of them at once. Proving a property for abstract vectors simultaneously establishes it for tuples, polynomials, matrices, and functions without repeating work.

## Practice questions

6 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. Polynomials of degree at most 2, 2x3 matrices, and arrows in the plane are studied as one subject. Why is it worth proving a theorem about "vector spaces" rather than proving it three times about three concrete objects?

A. Because the three objects are secretly the same object written differently, so the three proofs would be redundant.
B. Because the proofs would be identical, using only the shared properties.
C. Because abstract proofs are shorter than concrete ones, which saves space in textbooks.
D. Because concrete objects cannot be reasoned about rigorously, so abstraction is required for a valid proof.

**Answer:** B. Because the proofs would be identical, using only the shared properties.

**Why:** Try the experiment: prove the theorem for polynomials, then for matrices, and the two arguments turn out to use nothing about either, only that you can add, scale, and land back inside. The abstraction is the recognition that the proof was never about the objects. Solutions of a linear differential equation form a vector space, and every theorem was waiting for them.

Page: https://tryals.app/practice/mathematics-i/vector-spaces/polynomials-of-degree-at-most-2-2x3-matrices-and-arrows-in-the

### 2. The solutions of a linear differential equation form a vector space. Why is noticing that worth anything to someone trying to solve one?

A. Because it proves the equation has at least one solution
B. Because every theorem about vector spaces immediately applies
C. Because it means the solutions can be written as matrices
D. Because vector spaces are easier to compute with than functions

**Answer:** B. Because every theorem about vector spaces immediately applies

**Why:** Closure under sums and scaling means solutions have a basis, so a second-order equation needs exactly two independent ones and everything else follows as a combination. The whole solution method is a basis argument, which is the payoff for defining vector spaces abstractly instead of as arrows.

Page: https://tryals.app/practice/mathematics-i/vector-spaces/the-solutions-of-a-linear-differential-equation-form-a-vector-space

### 3. Select every set that IS a vector space under the usual operations.

A. All polynomials with a positive leading coefficient
B. All infinite sequences of real numbers
C. All 2x2 matrices whose entries sum to 0
D. All 2x2 matrices whose entries sum to 1
E. All polynomials p with p(1) = 0

**Answer:** B. All infinite sequences of real numbers; C. All 2x2 matrices whose entries sum to 0; E. All polynomials p with p(1) = 0

**Why:** Entry-sum 0 and p(1) = 0 are conditions that equal zero, so they are preserved by sums and scalings; sequences add and scale termwise. Entry-sum 1 excludes the zero matrix, and 'positive leading coefficient' is destroyed by multiplying by -1, and also excludes the zero polynomial.

Page: https://tryals.app/practice/mathematics-i/vector-spaces/select-every-set-that-is-a-vector-space-under-the-usual-operations

### 4. Order these vector spaces by how many real numbers it takes to specify one element, fewest first.

**Answer:**

1. The plane, that is all pairs (x, y)
2. Quadratics, written as a + bx + cx squared
3. 2x2 matrices
4. 3x2 matrices

**Why:** The counts are 2, 3, 4 and 6. Polynomials of degree at most 2 need three coefficients, so they need MORE numbers than the plane but fewer than a 2x2 matrix, the objects look nothing alike, yet this count is the only thing that distinguishes them structurally.

Page: https://tryals.app/practice/mathematics-i/vector-spaces/order-these-vector-spaces-by-how-many-real-numbers-it-takes-to

### 5. The set of all real-valued functions f with f(2) = 0 is a vector space under pointwise addition and scaling.

**Answer:** True

**Why:** True. If f(2) = 0 and g(2) = 0 then (f + g)(2) = 0 and (cf)(2) = 0, so the set is closed; the zero function belongs to it, and the remaining axioms are inherited from the space of all functions. Had the condition been f(2) = 1, closure would fail immediately.

Page: https://tryals.app/practice/mathematics-i/vector-spaces/the-set-of-all-real-valued-functions-f-with-f-2-0-is-a-vector

### 6. A vector space must be closed under addition and scaling and must contain a zero vector. Sort each set, taken with its usual operations.

**Answer:**

- Is a vector space: All polynomials of degree at most 3, All 2x2 matrices, All pairs (x, y) with x + y = 0, All functions from the reals to the reals
- Is not a vector space: All polynomials of degree exactly 3, All pairs (x, y) with x + y = 1

**Why:** Degree-exactly-3 polynomials fail twice: the zero polynomial is missing, and $x^{3}$ plus $(-x^{3} + x)$ has degree 1. The set x + y = 1 misses (0, 0). Everything else is closed under addition and scaling, so all the remaining axioms are inherited.

Page: https://tryals.app/practice/mathematics-i/vector-spaces/a-vector-space-must-be-closed-under-addition-and-scaling-and-must