Back to Linear independence

Exercise 8

Avg difficulty 2.5

When we explored matrix multiplication in Matrix multiplication and linear combinations, we saw that some properties that are true for real numbers are not true for matrices. This exercise will investigate that in some more depth.

  1. Suppose that AA and BB are two matrices and that AB=0AB = 0. If B≠0B \neq 0, what can you say about the linear independence of the columns of AA?

  2. Suppose that we have matrices AA, BB and CC such that AB=ACAB = AC. We have seen that we cannot generally conclude that B=CB=C. If we assume additionally that AA is a matrix whose columns are linearly independent, explain why B=CB = C. You may wish to begin by rewriting the equation AB=ACAB = AC as AB−AC=A(B−C)=0AB-AC = A(B-C) = 0.

Austin, Understanding Linear Algebra, §2.4 Linear independence, Exercise 8

AI Tutor

Adapted from Understanding Linear Algebra © 2023 David Austin. Licensed under CC BY 4.0. Source: Understanding Linear Algebra. Imported at revision 3e6dced.