# Subspaces

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

## What is a Subspace?

A **subspace** is a subset $W$ of a vector space $V$ that is **itself** a vector space under the same operations. Most action in linear algebra—solution sets, spans, kernels—happens inside subspaces.

Checking every axiom is tedious. Instead, use the **Subspace Test**: a nonempty subset $W$ is a subspace if it is **closed** under two operations:

| Operation | Rule | Meaning |
| --- | --- | --- |
| **Addition** | If $u, v \in W$, then $u + v \in W$ | Adding vectors stays inside |
| **Scaling** | If $v \in W$, $c$ is scalar, then $cv \in W$ | Scaling vectors stays inside |

Setting $c = 0$ reveals a vital shortcut: every subspace **must contain the zero vector** ($\mathbf{0}$). Any subset missing $\mathbf{0}$ fails instantly.

## Geometry of Subspaces

In $\mathbb{R}^3$, the only possible **subspaces** are strict geometric categories:

| Subspace Type | Description |
| --- | --- |
| **Zero subspace** | The origin point only ($\{\mathbf{0}\}$) |
| **Lines** | Any line **through the origin** |
| **Planes** | Any plane **through the origin** |
| **Full space** | All of $\mathbb{R}^3$ |

**Common pitfall:** A line or plane that is **offset** from the origin is **not** a subspace. It fails because it lacks $\mathbf{0}$, and scaling any of its points by $0$ lands at the origin, off the line.

Subspaces also include **homogeneous** solution sets $A\mathbf{x} = \mathbf{0}$ (the **null space**), spans, and linear map ranges.

## Practice questions

6 of this lesson's 13 practice questions, with answers. The full set is in the app.

### 1. Every subspace must contain the zero vector. Why does that single requirement rule out so many otherwise reasonable-looking sets?

A. Because it distinguishes finite from infinite subspaces
B. Because it excludes everything that misses the origin
C. Because the zero vector is the only element every subspace shares
D. Because sets without zero cannot be closed under addition

**Answer:** B. Because it excludes everything that misses the origin

**Why:** Almost every line and plane you can draw misses the origin, and one glance eliminates all of them. It is also not an independent axiom, scaling any element by 0 produces zero, so it is a fast necessary condition extracted from closure, which is exactly what makes it a good first filter.

Page: https://tryals.app/practice/mathematics-i/subspaces/every-subspace-must-contain-the-zero-vector-why-does-that-single

### 2. The subspace test asks for the zero vector first, before checking closure. Why is that the right order to apply the checks in, rather than a matter of taste?

A. Because it is the cheapest check and it disqualifies outright.
B. Because the zero vector is the only property that genuinely matters, and closure follows from it automatically.
C. Because the checks must be applied in the order the axioms were historically discovered.
D. Because a set can be closed under addition and scaling without containing zero, so the zero check catches cases closure would miss.

**Answer:** A. Because it is the cheapest check and it disqualifies outright.

**Why:** Both checks are necessary and cost very different amounts: the zero test is one evaluation, closure is a general argument about every pair. So the ordering is economics, run the cheap disqualifier first, and y = x + 1 is dismissed in a second. The test is not logically independent either: scaling any element by 0 produces zero, which is what makes it a good filter.

Page: https://tryals.app/practice/mathematics-i/subspaces/the-subspace-test-asks-for-the-zero-vector-first-before-checking

### 3. Order the steps of the subspace test applied to a subset W of a vector space V, cheapest check first.

**Answer:**

1. Check that W is nonempty by finding the zero vector in it
2. Take arbitrary u and v in W and check that u + v is still in W
3. Take an arbitrary v in W and scalar c and check that cv is still in W
4. Conclude that the remaining axioms are inherited from V, so W is a subspace

**Why:** The zero-vector check is free and rejects most impostors on sight. The two closure checks then do all the real work, and every other axiom comes free from the surrounding space V, which is why the test is only two conditions long.

Page: https://tryals.app/practice/mathematics-i/subspaces/order-the-steps-of-the-subspace-test-applied-to-a-subset-w-of-a

### 4. The 2x2 matrices with trace 0 (top-left entry plus bottom-right entry equal to zero) form a subspace of the 2x2 matrices. How many of the four entries can still be chosen freely?

**Answer:** 3

**Why:** The condition a + d = 0 means d = -a, so a, b and c are free and d follows: 3 free entries out of 4. One linear condition removes exactly one degree of freedom, the same accounting that will reappear as dimension in Lesson 5.

Page: https://tryals.app/practice/mathematics-i/subspaces/the-2x2-matrices-with-trace-0-top-left-entry-plus-bottom-right-entry

### 5. Select every subset of the plane that DOES contain the zero vector but is still NOT a subspace.

A. The union of the x-axis and the y-axis
B. All (x, y) with y equal to x cubed
C. The line y = 2x + 3
D. The closed first quadrant, where neither coordinate is ever negative
E. The line y = 2x

**Answer:** A. The union of the x-axis and the y-axis; B. All (x, y) with y equal to x cubed; D. The closed first quadrant, where neither coordinate is ever negative

**Why:** The two axes contain 0 and are closed under scaling, yet (1, 0) + (0, 1) = (1, 1) lies on neither axis. The first quadrant contains 0 and is closed under addition, but not under scaling by -1. The cubic contains 0 but (1, 1) + (2, 8) = (3, 9), and 9 is not 27. The line y = 2x is a genuine subspace, and y = 2x + 3 was excluded because it misses the origin.

Page: https://tryals.app/practice/mathematics-i/subspaces/select-every-subset-of-the-plane-that-does-contain-the-zero-vector

### 6. Complete the consequence of closure under scalar multiplication.

**Answer:** Because a nonempty subspace is closed under scaling by every scalar, taking c = 0 shows that it must **contain the zero vector**.

**Why:** Pick any v in W; closure gives 0v = 0 in W. This is why the zero-vector test is legitimate rather than an extra axiom, it is a consequence of closure, and it rejects every set offset from the origin in one line.

Page: https://tryals.app/practice/mathematics-i/subspaces/complete-the-consequence-of-closure-under-scalar-multiplication