After row-reducing a matrix A to RREF, a student uses the pivot columns of RREF as her column-space basis and the nonzero rows of RREF as her row-space basis. What is wrong with this procedure?
ABoth procedures are wrong — you must use the original matrix A for both bases
BThe column-space basis must come from the pivot columns of the ORIGINAL A, not the RREF; the row-space basis from the nonzero rows of RREF is correct
CBoth procedures are correct
DThe row-space basis must also come from the original matrix A, not the RREF
Row operations preserve the row space but change the column space. This creates an asymmetry: the nonzero rows of any row echelon form directly give a basis for the row space (because row operations produce only linear combinations of the original rows). But row operations alter the column directions, so you must identify pivot columns in the RREF to find which columns of the ORIGINAL A form a column-space basis. Using RREF columns for the column space is the classic error.
Question 2 Multiple Choice
A vector x satisfies Ax = 0 (i.e., x is in the null space of A). What is the geometric relationship between x and every vector in the row space of A?
Ax is parallel to every vector in the row space
Bx is orthogonal to every vector in the row space
Cx has the same dimension as the row space
Dx must be the zero vector
If Ax = 0, then every row of A dotted with x equals zero (since Ax computes exactly those dot products). This means x is perpendicular to every row vector of A, and therefore perpendicular to every linear combination of rows — i.e., every vector in the row space. The null space Nul(A) and the row space Row(A) are orthogonal complements in Rⁿ, with their dimensions summing to n by the rank-nullity theorem.
Question 3 True / False
Elementary row operations preserve the row space of a matrix but generally change its column space.
TTrue
FFalse
Answer: True
Each row operation (swap, scale, add a multiple of one row to another) produces rows that are linear combinations of the original rows. So the span of the rows — the row space — is unchanged. The column directions, however, are typically altered: adding a multiple of one row to another changes the entries in each column, so the column vectors in the RREF are generally different from those in the original A. This asymmetry is why the two bases are found by different procedures.
Question 4 True / False
For an m×n matrix A, the row space and the column space are subspaces of the same vector space.
TTrue
FFalse
Answer: False
The row space Row(A) is a subspace of Rⁿ (since each row has n entries), while the column space Col(A) is a subspace of Rᵐ (since each column has m entries). They live in different ambient spaces unless m = n. Despite this, they always have the same dimension — the rank of A — which is the surprising content of the theorem that row rank equals column rank.
Question 5 Short Answer
Explain why row rank equals column rank, and why this equality is surprising.
Think about your answer, then reveal below.
Model answer: The rank of A counts the number of nonzero rows in any echelon form, which equals the number of pivot positions. But the number of pivot positions also equals the number of pivot columns — hence the number of linearly independent columns. So row rank (dimension of the row space) and column rank (dimension of the column space) both equal the number of pivots. The surprise is that these two spaces live in different ambient spaces — Rⁿ and Rᵐ respectively — yet always have the same dimension. There is no obvious geometric reason why compressing m-dimensional column information and n-dimensional row information should yield the same count.
The proof goes through the fact that row reduction counts the same thing from both perspectives: the pivots mark both the independent rows and the independent columns simultaneously. The equality breaks symmetry in a deep way — it says that no matter how rectangular A is, its 'row complexity' and 'column complexity' are identical numbers, reflecting the same underlying rank.