Exercises: Vectors and Linear Combinations

Practice problems and worked solutions for vectors, span, dependence, and column combinations.

Keywords

exercises, vectors, span, linear combinations

1 Scope and Goals

This set trains four things:

• writing vectors explicitly as linear combinations
• checking span and dependence
• reading matrix-vector multiplication by columns
• preparing for basis and dimension

2 Prerequisites

• vector addition and scalar multiplication
• the definition of span
• the idea that \(Ax\) is a column combination

3 Warm-Up Problems

1. Write

   \[ \begin{bmatrix} 4 \\ -1 \end{bmatrix} \]

   as a linear combination of the standard basis vectors in \(\mathbb{R}^2\).

2. Let

   \[ v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \qquad v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}. \]

   Find coefficients \(a,b\) so that \(av_1+bv_2 = (3,1)^\top\).

4 Core Problems

1. Decide whether

   \[ b = \begin{bmatrix} 2 \\ 5 \end{bmatrix} \]

   belongs to the span of

   \[ v_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \qquad v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \]

2. Let

   \[ A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}, \qquad x = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}. \]

   Compute \(Ax\) and explain it as a linear combination of the columns of \(A\).

3. Show that

   \[ v_3 = \begin{bmatrix} 2 \\ 0 \end{bmatrix} \]

   is redundant in the set

   \[ \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \end{bmatrix} \right\}. \]

5 Proof Problems

1. Prove that if \(u\) and \(v\) are both in \(\operatorname{span}\{w_1,\dots,w_k\}\), then \(u+v\) is also in that span.
2. Prove that if one vector in a spanning set is itself a linear combination of the others, then removing it does not change the span.

6 Computational or Applied Problems

1. Open Computation Lab: Linear Combinations and Span Geometry. Move the angle close to 0 degrees and record what happens to the determinant and the span description.
2. In software, create a small matrix with dependent columns and test two different coefficient vectors that give the same output.

7 Hints

1. To check membership in a span, solve for the coefficients.
2. For matrix-vector multiplication, read each component of \(x\) as a weight on a column of \(A\).
3. Redundancy means one direction is already available from the others.

8 Full Solutions

8.1 Solution 1

\[ \begin{bmatrix} 4 \\ -1 \end{bmatrix} = 4 \begin{bmatrix} 1 \\ 0 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \end{bmatrix}. \]

8.2 Solution 2

We solve

\[ a \begin{bmatrix} 1 \\ 1 \end{bmatrix} + b \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}. \]

This gives

\[ a+b=3, \qquad a-b=1. \]

So \(a=2\) and \(b=1\).

8.3 Solution 3

Solve

\[ a \begin{bmatrix} 1 \\ 2 \end{bmatrix} + b \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \end{bmatrix}. \]

From the first coordinate, \(a+b=2\). From the second, \(2a+b=5\). Subtracting gives \(a=3\), so \(b=-1\).

Therefore \(b\) belongs to the span.

8.4 Solution 4

The columns of \(A\) are

\[ a_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad a_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \qquad a_3 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \]

So

\[ Ax = 2a_1 - a_2 + 3a_3 = 2 \begin{bmatrix} 1 \\ 0 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \end{bmatrix} + 3 \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \end{bmatrix}. \]

8.5 Solution 5

Because

\[ \begin{bmatrix} 2 \\ 0 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \]

the third vector is already in the span of the first. Removing it does not change the span.

9 Common Errors

• treating coefficients as if they must be unique even when the generators are dependent
• checking span by eye instead of solving for coefficients
• forgetting that matrix-vector multiplication is column-based, not row-based, in this interpretation

10 What To Do Next

Open Matrices and Linear Maps next. That page explains why these same linear combinations become operator language once matrices are viewed as maps.

11 Sources and Further Reading

• MIT 18.06SC: Basis and Dimension - First pass - good official problem context for span and dependence. Checked 2026-04-24.
• Hefferon, Linear Algebra - Second pass - strong extra exercises for span, dependence, and column space. Checked 2026-04-24.
• Attention is All you Need - Paper bridge - useful later, once weighted vector combinations feel natural. Checked 2026-04-24.