Linear Algebra, Live
Strang §1.3 and §3.1 · Column space

The column space C(A) is every possible Ax.

A matrix can only produce certain outputs. The column space is the list of everything Ax can produce. Solving Ax = b means asking one question: is b on that list?

C(A) = all vectors Ax = all combinations of the columns

Think: reachable outputs

The columns are the directions you are allowed to move in. The vector x says how far to go along each one. Ax is where you end up.

Try every possible x and mark every place you land. That set of landing spots is C(A).

An engineering picture

A small boat has two thrusters. Column 1 is the push you get from thruster 1 at full throttle. Column 2 is the push from thruster 2. The entries of x are the two throttle settings, and a negative throttle means reverse.

  • Thrusters at an angle: by mixing them you can push the boat in any direction. C(A) is the whole plane.
  • Both thrusters pointing forward: you can only go forward or back. C(A) is a line. A sideways target is out of reach, whatever the throttles.

Three possibilities in the plane

  • The whole plane: the columns point in different directions. They are independent.
  • A line: one column is a multiple of the other. They are dependent.
  • Just the origin: A is all zeros.

The rank counts how many dimensions survive: 2, 1 or 0. It is also the number of independent columns.

Worked example

A = 1224. Column 2 is 2 times column 1.
Ax = x₁(1, 2) + x₂(2, 4) = (x₁ + 2x₂)·(1, 2)
Every output is a multiple of (1, 2). So b = (3, 6) is reachable, and b = (1, 0) never is.
Notice that x = (3, 0) and x = (1, 1) both give (3, 6). With dependent columns, different inputs can land on the same output. Chapter 3 calls the leftover freedom the nullspace.

Two questions that look alike

  • Can I solve Ax = b for this b? Only if b is in C(A).
  • Can I solve Ax = b for every b? Only if C(A) is the whole plane.

For 2 by 2 there is a quick test. The columns are independent exactly when ad − bc ≠ 0. When it is 0, the plane gets squashed onto a line. A squash cannot be undone, so A has no inverse.

Worked examples

Example 1 · A line, and who is on it

A = 2412. Column 2 = 2·column 1, so C(A) is the line through (2, 1).

b = (6, 3) = 3·(2, 1): in C(A), for example x = (3, 0)
b = (1, 1): not a multiple of (2, 1), so Ax = b has no solution

Example 2 · The whole plane

A = 111−1. det A = −1 − 1 = −2 ≠ 0, so C(A) is the whole plane. Solve Ax = (4, 2):

x₁ + x₂ = 4    x₁ − x₂ = 2
x₁ = 3, x₂ = 1

Problems

Try each one before you open the solution.

1

Describe C(A) for A = 3612. What is the rank?

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Column 2 = 2·column 1, so C(A) is the line through (3, 1). Rank 1.

2

For which t is (t, 4) in the column space of 1224?

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C(A) is the line of multiples of (1, 2). (t, 4) = 2·(1, 2) needs t = 2.

3

What is C(A) when A is the zero matrix? When A is the identity?

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Zero matrix: only the origin, rank 0. Identity: the whole plane, rank 2, since Ix = x reaches everything.

Exercise Change column 2 into a multiple of column 1 and watch C(A) shrink to a line. Edit the code and press Run.
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✎Lesson review
3 more worked examples and 5 mixed problems with solutions
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