Questions: Column Space and Row Space

Question 1

The system Ax = b has no solution. What does this tell you about b in relation to A?

• A. b is in the null space of A
• B. b is not in the column space of A
• C. b is a linear combination of the rows of A
• D. b equals the zero vector

Answer: (B)

Ax = b has a solution if and only if b is in the column space of A. Since Ax computes a linear combination of A's columns with x as coefficients, the set of all achievable right-hand sides is exactly col(A). If b is not in col(A), no combination of the columns can produce it, so no solution exists. Option A (null space) is the set of x where Ax = 0 — it describes inputs that map to zero, not right-hand sides.

Question 2

A student row-reduces a matrix A to its echelon form and finds that columns 1 and 3 are pivot columns. To find a basis for col(A), she takes columns 1 and 3 from the echelon form. What has she done wrong?

• A. Nothing — the echelon form's pivot columns form a valid basis for col(A)
• B. She should take columns 1 and 3 from the original matrix A, not the row-reduced form
• C. She should use the rows rather than the columns to find the column space basis
• D. She should take all columns, not just the pivot ones

Answer: (B)

Row reduction identifies *which* columns are pivot columns, but the basis vectors for col(A) must be taken from the *original* matrix A, not the row-reduced form. Row operations preserve the row space but change the column space — the specific column vectors change. The correct procedure: row-reduce to identify pivot positions, then return to A and take those columns as the basis.

Question 3

Row reduction preserves the column space of a matrix.

• T. True
• F. False

Answer: False

Row reduction preserves the *row space* but generally changes the *column space*. The column vectors change when you perform elementary row operations. This is why you must identify pivot columns from the row-reduced form but return to the original matrix to extract the actual basis vectors. The row and column spaces have the same dimension (rank), but they are found by different procedures.

Question 4

The column space of an m × n matrix A is a subspace of ℝⁿ.

• T. True
• F. False

Answer: False

The column space of an m × n matrix A is a subspace of ℝᵐ, not ℝⁿ. The columns of A are vectors in ℝᵐ (each column has m entries), so their span lives in ℝᵐ. The row space, by contrast, is a subspace of ℝⁿ. It's easy to confuse these because m and n both appear in descriptions of A — keeping track of which dimension corresponds to rows versus columns is essential.

Question 5

Why does the equation Ax = b have a solution if and only if b is in the column space of A? Explain in terms of what matrix-vector multiplication actually computes.

Model answer: Matrix-vector multiplication Ax computes a linear combination of the columns of A, where the entries of x are the scalar coefficients. If A = [a₁ | a₂ | ... | aₙ], then Ax = x₁a₁ + x₂a₂ + ... + xₙaₙ. The set of all such linear combinations, as x ranges over all of ℝⁿ, is exactly col(A). So Ax = b asks: can b be written as a linear combination of A's columns? If yes, the corresponding coefficients are a solution x. If no, b is outside the reachable set, and no solution exists.

This interpretation is the geometric heart of linear algebra. The column space is the image of the transformation T(x) = Ax — the set of all outputs the transformation can produce. Understanding Ax as a linear combination of columns (rather than mechanically multiplying rows by x) unlocks the connection between solvability of Ax = b and the geometry of col(A).