Exercise 2.3.21. Suppose is a 64 by 17 matrix and has rank 11. How many independent vectors are solutions to the system ? What about the system ?

Answer: If the rank of is 11 then performing elimination on produces an matrix with 11 pivots and thus 11 basic variables. Since (like ) has 17 columns this means that there are 17 – 11 or 6 free variables that can be set to arbitrary values in solving the system . The nullspace of (i.e., the set of all vectors satisfying ) therefore has dimension 6, and any basis for the nullspace has 6 linearly independent vectors each of which satisfy .

Since is 64 by 17 the matrix is 17 by 64. The original matrix had 11 pivots and 11 linearly independent rows. The rows of become columns in and thus has 11 linearly independent columns and also has rank 11. Since has 64 columns there are 64 – 11 or 53 free variables when considering the system . The nullspace of (i.e., the set of all vectors satisfying ) therefore has dimension 53, and any basis for the nullspace has 53 linearly independent vectors each of which satisfy .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.