Exercise 2.4.11. Suppose that for a matrix and any the system always has at least one solution. Show that in this case the system has no solution other than .

Answer: Suppose is an by matrix and is an -element vector. If is a solution for then is a linear combination of the columns of with the coefficients being .

If we can find a solution for any in then the columns of span since any vector in can be expressed as a linear combination of the columns of . The rank of is therefore . (The rank cannot be less than since then there would not be enough linearly independent columns of to span , and cannot be greater than because it is impossible to have more than linearly independent vectors in .)

Now consider the left nullspace consisting of all vectors such that . The dimension of the left nullspace is where is the rank of . But from above we have so that the dimension of is . The only element of is therefore the zero vector, and there are no solutions to other than .

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.