Dimension and the Rank-Nullity Theorem · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Rank plus nullity equals columns

Mathematics · Linear Algebra · ages 18-19
Name ______________________   Date ____________

  1. A consistent system Ax = b has a 4 by 6 coefficient matrix of rank 3. How many free parameters does its solution set have?

    • 2
    • 4
    • 6
    • 3

  2. A 6 by 8 matrix has rank 5. What is its nullity?

    Answer: ______________

  3. A 4 by 5 matrix has rank 3. What is its nullity?

    Answer: ______________

  4. A null space is spanned by (1, 0, -1) and (0, 1, 2). Neither is a multiple of the other. What is the nullity?

    Answer: ______________

  5. Tom says the column space of [[1, 2], [2, 4]] is all of R squared. Is Tom right?

    Circle one:   True   False

  6. A 3 by 4 matrix can have nullity 0.

    Circle one:   True   False

  7. A has columns (1, 0, 1), (2, 0, 2) and (0, 1, 1), and its reduced form has pivots in columns 1 and 3. Which set is a basis for the column space?

    • (1, 0, 1) and (0, 1, 1)
    • (1, 0, 1) and (2, 0, 2)
    • (1, 0, 0) and (0, 1, 0)
    • (2, 0, 2) alone

  8. A 6 by 6 matrix has rank 6. What does that say about Ax equals b?

    • It has exactly one solution for every b
    • It has infinitely many solutions for every b
    • It has no solution for some b

  9. Can a 4 by 7 matrix have rank 5?

    • Yes, rank 5 is possible here
    • Yes, rank 5 is forced for wide matrices
    • No, the rank cannot pass the 4 rows

  10. A matrix has 5 columns and nullity 3. How many vectors are in a basis for its column space?

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

Rank plus nullity equals columns W1-mt_aMG8PcwPAR-s1

  1. 3 · The 6 unknowns minus rank 3 leave nullity 3, so there are 3 free parameters.
  2. 3 · Rank plus nullity equals the 8 columns, so nullity is 8 minus 5, which is 3.
  3. 2 · Rank plus nullity equals the 5 columns, so nullity is 5 minus 3.
  4. 2 · The two spanning vectors are independent, so they form a basis of size 2.
  5. False · Tom is wrong: the second column is twice the first, so the column space is just a line.
  6. False · Rank is at most 3, so nullity is at least 4 minus 3, which is 1.
  7. (1, 0, 1) and (0, 1, 1) · Keep the original columns in the pivot positions, columns 1 and 3. (2, 0, 2) is twice (1, 0, 1), and (1, 0, 0) is not in the column space.
  8. It has exactly one solution for every b · Full rank means nullity 0, so solutions are unique, and 6 pivots reach every b.
  9. No, the rank cannot pass the 4 rows · Rank is bounded by rows and columns alike, so at most 4 here.
  10. 2 · Rank is 5 minus 3, which is 2, and the rank is the size of a column space basis.
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