# Vector Spaces

Business I · Mathematics I · https://tryals.app/learn/business-i/vector-spaces

## The Anatomy of Vector Spaces

A **vector** is a list of numbers representing data like asset weights or sales. A **vector space** is the universe built from vectors using two moves: addition and scaling.

**Linear combination**: the core building move, written as $w = \lambda_1 v_1 + \dots + \lambda_k v_k$. Mixing stocks and bonds in any proportions creates a linear combination. The set of all such combinations is their **span**, the entire reachable territory.

| Concept | Definition | Practical Test |
| --- | --- | --- |
| **Span** | All reachable linear combinations | Territory you can build |
| **Linear Independence** | No vector is a combination of others | No redundancy |
| **Basis** | Independent and spanning set | Minimal coordinate system |

**Linear independence** means no vector is redundant. If fund C is a 50/50 split of funds A and B, adding it **spans nothing new**.

## Basis, Dimension, and Variety

A **basis** is a set that is both independent (no waste) and spanning (no gaps). **Dimension** is the number of vectors in any basis, counting genuine degrees of freedom.

| Property | Meaning |
| --- | --- |
| **Dimension** | Count of independent directions, not listed items |
| **Matrix Rank** | Number of independent vectors when stacked as columns |
| **Determinant** | $\det \neq 0$ for $n$ independent vectors in $\mathbb{R}^n$ |

**Common pitfall:** Confusing the number of listed items with the true dimension. Three portfolios built from just two funds still span a 2D space. Diversifying across dependent assets is buying identical risk in different wrappers. 

**Tip:** To test independence, stack vectors as columns and compute the rank. Rank equals the vector count if and only if they are independent.

## Practice questions

8 of this lesson's 14 practice questions, with answers. The full set is in the app.

### 1. Order the reasoning to find the dimension of the space spanned by a set of vectors.

**Answer:**

1. Stack the vectors as rows of a matrix
2. Run Gaussian elimination
3. Count the nonzero rows (the rank)
4. Discard the redundant vectors
5. The rank is the dimension of the span

**Why:** Stack, eliminate, count, discard, conclude. Gaussian elimination is a redundancy detector: every zero row it produces was a vector pretending to be new.

Page: https://tryals.app/practice/business-i/vector-spaces/order-the-reasoning-to-find-the-dimension-of-the-space-spanned-by-a

### 2. Sort each investment-menu situation by its linear-algebra diagnosis.

**Answer:**

- Adds a new dimension: A commodity fund uncorrelated with your stock/bond mix, Real estate exposure your portfolio cannot replicate
- Redundant (dependent): Fund C is always exactly 50% A plus 50% B, A leveraged fund that is 2× an index you already hold, An "aggressive" fund = 80% fund A + 20% fund B

**Why:** A product diversifies only if it is *linearly independent* of what you hold. Leverage and remixes are old vectors with new labels, span unchanged.

Page: https://tryals.app/practice/business-i/vector-spaces/sort-each-investment-menu-situation-by-its-linear-algebra-diagnosis

### 3. Match each concept to its definition.

**Answer:**

- Linear combination → Scaled vectors added together
- Span → Everything reachable by combinations
- Basis → Independent set that spans the space
- Dimension → Number of vectors in any basis

**Why:** Build, reach, coordinate, count, the four-word ladder of vector-space theory.

Page: https://tryals.app/practice/business-i/vector-spaces/match-each-concept-to-its-definition

### 4. Why must every basis of $\mathbb{R}^{3}$ have exactly three vectors, never two, never four?

A. Because fewer cannot span and more must be dependent
B. Because four vectors would span a four-dimensional space
C. Because three is the number of coordinate axes, which is a convention
D. Because a basis is defined as a set of exactly three vectors

**Answer:** A. Because fewer cannot span and more must be dependent

**Why:** Two requirements squeeze from opposite sides and meet at the dimension. This is why dimension is well defined at all: every basis is forced to the same size.

Page: https://tryals.app/practice/business-i/vector-spaces/why-must-every-basis-of-have-exactly-three-vectors-never-two

### 5. An analyst tracks 12 macroeconomic series and finds they are all built from 3 underlying factors. What is that finding in vector-space language?

A. The 12 vectors span a 3-dimensional subspace
B. The data has rank 12 but only 3 non-zero components
C. The 12 series are linearly independent within a 3-dimensional space
D. Three of the series are correct and the other nine are measurement noise

**Answer:** A. The 12 vectors span a 3-dimensional subspace

**Why:** Dimension counts genuinely independent directions rather than columns of data, so twelve series can carry three series' worth of information. Options 2 and 3 are self-contradictory, you cannot have twelve independent vectors in a three-dimensional space, and this is exactly what factor models exploit.

Page: https://tryals.app/practice/business-i/vector-spaces/an-analyst-tracks-12-macroeconomic-series-and-finds-they-are-all

### 6. Which statements about span and independence are true?

A. In a basis, every vector of the space has unique coordinates
B. The zero vector can belong to an independent set
C. Adding a dependent vector to a set leaves the span unchanged
D. Two independent vectors in $\mathbb{R}^2$ span the whole plane

**Answer:** A. In a basis, every vector of the space has unique coordinates; C. Adding a dependent vector to a set leaves the span unchanged; D. Two independent vectors in $\mathbb{R}^2$ span the whole plane

**Why:** The zero vector is everyone’s dependent friend ($1\cdot 0 = 0$ is a nonzero-coefficient combination giving zero), so it poisons independence instantly. The rest are the core theorems.

Page: https://tryals.app/practice/business-i/vector-spaces/which-statements-about-span-and-independence-are-true

### 7. $(2,2)$ and $(3,3)$ together span only the line $y = x$, while $(1,0)$ and $(0,1)$ span the whole plane. What does that show about counting vectors?

A. That two vectors always span a line and three are needed for a plane
B. That spanning depends on whether the vectors are unit vectors
C. That the number of vectors does not determine the dimension
D. That $(2,2)$ and $(3,3)$ are too similar in magnitude to span a plane

**Answer:** C. That the number of vectors does not determine the dimension

**Why:** A dependent vector adds no new direction. This is exactly why rank is defined as the number of independent columns rather than the number of columns.

Page: https://tryals.app/practice/business-i/vector-spaces/2-2-and-3-3-together-span-only-the-line-y-x-while-1-0-and

### 8. Four vectors in $\mathbb{R}^3$ can be linearly independent.

**Answer:** False

**Why:** A 3-dimensional space has room for at most 3 independent directions; a fourth vector is always a combination of a basis. More vectors than dimensions forces dependence, no exceptions.

Page: https://tryals.app/practice/business-i/vector-spaces/four-vectors-in-can-be-linearly-independent