Exercises: Subspaces, Basis, and Dimension

Practice problems and worked solutions for subspace checks, basis construction, and dimension counting.

Keywords

exercises, subspace, basis, dimension

1 Scope and Goals

This set trains four things:

• checking subspace conditions carefully
• extracting a basis from a spanning description
• counting dimension correctly
• connecting column space language to later least-squares geometry

2 Prerequisites

• span and linear combinations
• independence
• column space as a set of outputs \(Ax\)

3 Warm-Up Problems

1. Decide whether

   \[ \left\{ \begin{bmatrix} x \\ y \\ z \end{bmatrix} : x+y+z=0 \right\} \]

   is a subspace of \(\mathbb{R}^3\).

2. Find a basis for the line

   \[ \operatorname{span}\left\{ \begin{bmatrix} 2 \\ -1 \end{bmatrix} \right\}. \]

4 Core Problems

1. Find a basis and the dimension of

   \[ W = \operatorname{span} \left\{ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \right\}. \]

2. Explain why

   \[ \left\{ \begin{bmatrix} x \\ y \end{bmatrix} : x+y=1 \right\} \]

   is not a subspace.

3. Let

   \[ A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}. \]

   Describe a basis for \(\operatorname{col}(A)\).

5 Proof Problems

1. Show that the column space of any matrix is a subspace.
2. Prove that removing a redundant vector from a spanning set does not change the span.

6 Computational or Applied Problems

1. Open Computation Lab: Basis and Column Space Geometry. Change the angle and explain when the third vector changes the subspace and when it does not.
2. In software, create a matrix with three columns in \(\mathbb{R}^2\) and compare the number of columns with the actual dimension of the column space.

7 Hints

1. A subspace must contain the zero vector.
2. To test independence, solve \(a_1v_1 + a_2v_2 = 0\).
3. Different spanning sets can define the same subspace.

8 Full Solutions

8.1 Solution 1

Yes. The set contains \(0\), and if \(x+y+z=0\) and \(x'+y'+z'=0\), then

\[ (x+x') + (y+y') + (z+z') = 0. \]

Scalar multiples also preserve the equation, so the set is a subspace.

8.2 Solution 2

A basis is just

\[ \left\{ \begin{bmatrix} 2 \\ -1 \end{bmatrix} \right\}. \]

The dimension is \(1\).

8.3 Solution 3

The two given vectors are independent, so they already form a basis. Therefore

\[ \dim(W)=2. \]

8.4 Solution 4

The vector \((0,0)^\top\) is not in the set because \(0+0 \neq 1\). So the set cannot be a subspace.

8.5 Solution 5

The columns are

\[ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \]

The first two already span \(\mathbb{R}^2\), and the third is their sum. So a basis is the first two columns.

9 Common Errors

• forgetting that a subspace must contain zero
• confusing ambient dimension with subspace dimension
• thinking every spanning set is already a basis

10 What To Do Next

Open Orthogonality and Least Squares next. That page turns column space into the target subspace for approximation.

11 Sources and Further Reading

• MIT 18.06SC: Basis and Dimension - First pass - official exercise context for span, basis, and dimension. Checked 2026-04-24.
• Hefferon, Linear Algebra - Second pass - stronger self-study exercise source on subspaces and bases. Checked 2026-04-24.
• A Survey: Potential Dimensionality Reduction Methods - Paper bridge - later reminder that low-dimensional models need honest basis and dimension reasoning. Checked 2026-04-24.