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

Rank is field independent

Let B be a 3×5 matrix with entries from . Assume that { | Bv = 0} is a three-dimensional real vector space. Which of the following statements are true?

  1. A.{ | Bv = 0} is a three-dimensional vector space over .
  2. B.The linear transformation given by T(v) = Bᵗv is injective.
  3. C.The column span of B is two-dimensional.
  4. D.The linear transformation given by T(v) = BBᵗv is injective.

Solution

Rank does not change under field extension, so rank B = 5 − 3 = 2 over as well: nullity over is 3 and the column span is 2-dimensional. Bᵗ : has rank 2 < 3, not injective; BBᵗ has rank 2 < 3, not injective.

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

From Vector Spaces and Linear MapsBases, dimension, rank–nullity

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