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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.
Exercise 2.3.20.Consider the set of all 2 by 2 matrices that have the sum of their rows equal to the sum of their columns. What is a basis for this subspace? Consider the analogous set of 3 by 3 matrices with equal row and column sums. List five linearly independent matrices from this set.
Answer: Any 2 by 2 matrix in the set will have the form
with the sum of every row and every column being . Any such matrix can be represented as a linear combination of two matrices as follows:
Since the two matrices are linearly independent and span the subspace they are a basis for the subspace.
The following five matrices are linearly independent members of the analogous set for 3 by 3 matrices:
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.
Exercise 2.3.19. Suppose is an by matrix, with columns taken from . What is the rank of if its column vectors are linearly independent? What is the rank of if its column vectors span ? What is the rank of if its column vectors are a basis for ?
Answer: If the column vectors of are linearly independent then there must be a pivot in every one of the columns, so that the rank .
If the columns of span then we must have . There can be no more than linearly independent vectors in so out of the columns of only columns can have pivots. Therefore the rank .
If the columns of are a basis for then they are linearly independent, which means the rank , and they also span so we must also have . We thus have .
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.
