Linear Algebra
5.610 min read

Linear Systems Revisited via Row Rank

With the Row Rank Theorem in hand, we can revisit solvability from a row perspective. Think of each equation in the system Ax=bA\mathbf{x} = \mathbf{b} as one row of AA paired with one entry of b\mathbf{b}. Adding an equation is adding a row.

The system is always solvable (for every possible b\mathbf{b}) if and only if Rank(A)=m\text{Rank}(A) = m. By the Row Rank Theorem, this means the row vectors of AA are linearly independent — no equation is a redundant combination of others.

Similarly, the dimension of the solution set is n−Rank(A)n - \text{Rank}(A). Using the Row Rank Theorem, this equals n−Rank(At)n - \text{Rank}(A^t). We started in Rn\mathbb{R}^n and each new linearly-independent row reduces the solution dimension by 1, exactly as our geometric intuition from Chapter 1 suggested.

Formal View

Theorem 5.11 — Always Solvable via Row Rank
Let AA be an m×nm \times n matrix. Then Ax=bA\mathbf{x} = \mathbf{b} is solvable for every b∈Rm\mathbf{b} \in \mathbb{R}^m if and only if the row vectors of AA are linearly independent.

Proof: always solvable iff Rank(A)=m\text{Rank}(A) = m. By the Row Rank Theorem, Rank(A)=Rank(At)=m\text{Rank}(A) = \text{Rank}(A^t) = m iff the columns of AtA^t (= rows of AA) are linearly independent.

Theorem 5.12 — Solution Dimension via Row Rank
If Ax=bA\mathbf{x} = \mathbf{b} is solvable, its solution set is an affine space of dimension
n−Rank(A)=n−Rank(At)=n−Rowrank(A).n - \text{Rank}(A) = n - \text{Rank}(A^t) = n - \text{Rowrank}(A).
Each linearly independent row reduces the solution dimension by exactly 11.

Why This Matters

Understanding solvability through rows gives direct geometric insight — each equation is a hyperplane constraint.

  • Network flow problems: each conservation law is a row; the system is solvable iff the laws are independent
  • Overdetermined systems (m>nm > n) almost never have exact solutions — motivating least squares regression
  • Circuit analysis (Kirchhoff's laws) gives one equation per node/loop; linear independence of those equations determines the circuit's degrees of freedom

Quiz

Question 1

A 3×53 \times 5 matrix AA has linearly independent rows. For how many right-hand sides b∈R3\mathbf{b} \in \mathbb{R}^3 is Ax=bA\mathbf{x} = \mathbf{b} solvable?

Question 2

If AA is 3×53 \times 5 with Rank(A)=3\text{Rank}(A) = 3, what is the dimension of the solution set of Ax=bA\mathbf{x} = \mathbf{b} (when it exists)?

Common Mistakes

  • Thinking more rows always means a harder system — only linearly independent rows reduce the solution space.
  • Confusing the condition for a unique solution (Rank(A)=n\text{Rank}(A) = n) with the condition for always solvable (Rank(A)=m\text{Rank}(A) = m).
  • Forgetting that n−Rank(A)n - \text{Rank}(A) counts free variables even in inhomogeneous systems — the solution set has the same dimension as the null space.