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Part CCSIR NET June 2023rank-is-field-independent

Rank is field independent

Let B be a 3×5 matrix with entries from Q\mathbb{Q}. Assume that {v∈R5v \in \mathbb{R}^{5} | Bv = 0} is a three-dimensional real vector space. Which of the following statements are true?

  1. A.{v∈Q5v \in \mathbb{Q}^{5} | Bv = 0} is a three-dimensional vector space over Q\mathbb{Q}.✓ Correct
  2. B.The linear transformation T:Q3→Q5T : \mathbb{Q}^{3} \to \mathbb{Q}^{5} given by T(v) = Bᵗv is injective.
  3. C.The column span of B is two-dimensional.✓ Correct
  4. D.The linear transformation T:Q3→Q3T : \mathbb{Q}^{3} \to \mathbb{Q}^{3} given by T(v) = BBᵗv is injective.

The idea, in plain words

Rank is computed by row reduction, and row reduction never leaves the field the entries came from. So enlarging the field cannot change the rank — the dimension counts are the same whether you work over the rationals or the reals.

Solution

Rank does not change under field extension

so rank B = 5 − 3 = 2 over Q\mathbb{Q} as well: nullity over Q\mathbb{Q} is 3 and the column span is 2-dimensional.

Bᵗ : Q3→Q5\mathbb{Q}^{3} \to \mathbb{Q}^{5} has rank 2 < 3, not injective; BBᵗ has rank 2 < 3, not injective.

Why each option is right or wrong

Checked against 2 symbolic computations
  • A.Correct

    True, and the key fact is that rank does not change under field extension — row reduction uses only field operations, and B has rational entries

    so the same pivots appear over Q\mathbb{Q} and over R\mathbb{R}.

    The real nullity is 3

    so rank B = 5 − 3 = 2 over both fields, and the rational null space is also 5 − 2 = 3 dimensional.

    The trigger: for a matrix over a subfield, rank, nullity and row space dimension are all field-independent.

  • B.WrongHypothesis dropped

    False.

    Bᵀ is 5 × 3 with rank 2 — SymPy confirms rank 2 on a representative rank-2 matrix of this shape — so the map Q3→Q5\mathbb{Q}^{3} \to \mathbb{Q}^{5} has a one-dimensional kernel and is not injective.

    A tall matrix is injective only when its rank equals its number of columns

    which is 3 here, not 2.

  • C.Correct

    True.

    The column span is the image of B, whose dimension is rank B = 2 (verified).

    Note this is the column span in Q3\mathbb{Q}^{3}

    sitting as a plane inside three-dimensional space — which is also why (D) fails.

  • D.WrongExecution slip

    False. rank(BBᵀ) = rank(B) = 2 for a real (or rational) matrix — SymPy confirms rank 2 for the corresponding 3 × 3 product.

    A map Q3→Q3\mathbb{Q}^{3} \to \mathbb{Q}^{3} of rank 2 has a non-trivial kernel

    so it is not injective.

    Multiplying by the transpose cannot raise the rank.

2 of these 2 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Base field or ring

The answer changes with the field or ring you are working over.

Drill statements like this

Related counterexample: An injective linear operator on a vector space is surjective

More on this topic

The chapter behind this: Bases, dimension and rank — free to read

From Vector Spaces and Linear Maps › Bases, dimension, rank–nullity

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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