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Why does the equation Ax=b have no solution when b is not in the column space C(A)C(A)?

The equation Ax⃗=b⃗A\vec{x}=\vec{b} has no solution because solving it is equivalent to finding weights x1,…,xkx_1, \dots, x_k such that x1a⃗1+⋯+xka⃗k=b⃗x_1\vec{a}_1 + \dots + x_k\vec{a}_k = \vec{b}. The column space C(A)C(A) is defined as the set of all possible linear combinations of the columns of AA. If b⃗\vec{b} lies outside this subspace, no combination of the columns can produce it, making the system inconsistent.

Conditions

  • AA is an n×kn \times k matrix
  • x⃗∈Rk\vec{x} \in \mathbb{R}^k and b⃗∈Rn\vec{b} \in \mathbb{R}^n
  • C(A)C(A) denotes the column space of AA

Reasoning, step by step

  1. Rewrite the matrix equation Ax⃗=b⃗A\vec{x}=\vec{b} as a linear combination of the columns of AA: x1a⃗1+x2a⃗2+⋯+xka⃗k=b⃗x_1\vec{a}_1 + x_2\vec{a}_2 + \cdots + x_k\vec{a}_k = \vec{b}.
  2. Identify that the left-hand side represents an arbitrary vector in the column space C(A)C(A).
  3. Observe the geometric condition that b⃗\vec{b} is not in C(A)C(A).
  4. Conclude that since b⃗\vec{b} cannot be formed by any linear combination of the columns, there are no weights xix_i that satisfy the equation.

Example

The video draws a purple plane representing C(A)C(A) and a cyan vector b⃗\vec{b} pointing outside that plane, visually demonstrating that b⃗\vec{b} cannot be reached by any combination of the columns.

Common misconceptions

  • Thinking that row reduction contradictions like 0=10=1 mean the problem is unsolvable in any sense, ignoring the possibility of approximation.
  • Confusing the column space with the row space or null space.

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.