# Basis and Dimension

Mathematics I · Linear Algebra · https://tryals.app/learn/mathematics-i/basis-and-dimension

## The Coordinate Skeleton

A **basis** for a vector space $V$ is a set of vectors that is **linearly independent** (no redundancy) and **spanning** (reaches every vector in $V$). It is a **minimal spanning set**: just enough vectors to build everything.

| Concept | Definition | Rule |
| --- | --- | --- |
| **Basis** | Independent + Spanning | Minimal spanning set |
| **Standard Basis** | $\mathbb{R}^n$ unit vectors | $e_1 = (1,0,\dots)$ |
| **Dimension** $\dim V$ | Size of any basis | Fundamental invariant |

Every basis of a given space has the **exact same number of vectors**. That number is the **dimension** $\dim V$ (e.g., $\mathbb{R}^2$ is 2, $\mathbb{R}^3$ is 3).

## Coordinates & Shortcuts

Because a basis spans, every vector is a combination of basis vectors; because it is independent, that combination is **unique**. This gives every vector a unique list of **coordinates**.

| Test in $\dim n$ Space | Result | Pitfall |
| --- | --- | --- |
| $n$ independent vectors | Automatically a basis | Don't assume spanning |
| $n$ spanning vectors | Automatically a basis | Don't assume independence |

> **Common pitfall:** Thinking any spanning set or independent set alone is a basis. A basis requires **both** properties, and always contains **exactly** $\dim V$ vectors.

## Practice questions

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

### 1. V is a vector space of dimension 4. Select every statement that MUST be true.

A. Any 3 vectors in V span V
B. Any 5 vectors in V are linearly dependent
C. Every basis of V has exactly 4 vectors
D. Any 4 linearly independent vectors in V form a basis
E. Any 4 vectors in V form a basis

**Answer:** B. Any 5 vectors in V are linearly dependent; C. Every basis of V has exactly 4 vectors; D. Any 4 linearly independent vectors in V form a basis

**Why:** All bases share the dimension count, and in an n-dimensional space n independent vectors are automatically spanning. But four arbitrary vectors could be four copies of one vector, and three vectors can never reach a four-dimensional space. Five vectors exceed the dimension and must be dependent.

Page: https://tryals.app/practice/mathematics-i/basis-and-dimension/v-is-a-vector-space-of-dimension-4-select-every-statement-that-must

### 2. A JPEG discards most of the numbers describing an image and it still looks right. Which linear-algebra idea is being used?

A. A change of basis, where most coefficients fall near zero
B. Rounding each pixel value to fewer decimal places
C. Dimension reduction, which lowers the image's dimension below what the eye can detect
D. Rank deficiency, since photographs are rarely full rank

**Answer:** A. A change of basis, where most coefficients fall near zero

**Why:** Same space, different basis. Written in pixels every number is essential; written in frequencies most coefficients are near zero, so throwing them away costs little. Choosing a basis in which your data is sparse is the whole of transform compression, and the same move drives PCA and wavelets.

Page: https://tryals.app/practice/mathematics-i/basis-and-dimension/a-jpeg-discards-most-of-the-numbers-describing-an-image-and-it-still

### 3. Order these subspaces of ordinary three-dimensional space by increasing dimension.

**Answer:**

1. The subspace containing only the origin
2. The span of (1, 2, 2)
3. The plane x + y + z = 0
4. The span of (1,0,0), (0,1,0) and (0,0,1)

**Why:** The dimensions are 0, 1, 2 and 3. The plane x + y + z = 0 is cut out by one equation, so two of the three coordinates stay free, each independent linear condition costs exactly one dimension.

Page: https://tryals.app/practice/mathematics-i/basis-and-dimension/order-these-subspaces-of-ordinary-three-dimensional-space-by

### 4. Because $\{1, x, x^{2}\}$ is a basis for polynomials of degree at most 2, every such polynomial has exactly one expression in terms of it.

**Answer:** True

**Why:** True. Spanning gives at least one expression; independence gives at most one, and together exactly one. That uniqueness is what makes coordinates meaningful, it is why $ax^{2}+bx+c$ can be identified with the triple $(c,b,a)$ and why the space is 3-dimensional.

Page: https://tryals.app/practice/mathematics-i/basis-and-dimension/because-1-x-x-is-a-basis-for-polynomials-of-degree-at-most-2

### 5. A basis is required to be both spanning and independent. Why does dropping either requirement break the thing a basis is for - giving every vector exactly one set of coordinates?

A. Because independence alone is sufficient, and the spanning condition is a historical redundancy.
B. Because a basis must be as small as possible, and the two conditions are two ways of saying that.
C. Because coordinates require the basis vectors to be orthogonal, which follows from the two conditions together.
D. Because spanning gives coordinates and independence makes them unique.

**Answer:** D. Because spanning gives coordinates and independence makes them unique.

**Why:** The two conditions do different jobs and each failure has its own symptom. Drop spanning and (0,1) cannot be written in terms of (1,0) at all. Drop independence and (2,1) has infinitely many representations over {(1,0), (0,1), (1,1)}. Together they give existence and uniqueness, which is what lets an abstract n-dimensional space be treated as a list of n numbers.

Page: https://tryals.app/practice/mathematics-i/basis-and-dimension/a-basis-is-required-to-be-both-spanning-and-independent-why-does