Testing for Independence

Linear Algebra (Matrices) · Axiom Academy

EXAMPLE Testing for Independence Make the vectors the columns of a matrix, row-reduce, and read off the verdict from the pivots Decide whether the three vectors in below are linearly independent . They are independent exactly when forces c_1=c_2=c_3=0 — i.e. when has only the trivial solution, where A has these vectors as its columns. Nice work — you tested a set of vectors for independence with one row reduction. The pieces worth keeping: The test: are independent precisely when (columns of A are the vectors) has only the trivial solution . Read it from the pivots: a pivot in every column means no free variables, so the only solution is trivial — independent . A free variable means dependence: any pivot-free column gives a free variable, hence a nontrivial , and one vector is a combination of the others — dependent . Homogeneous shortcut: the augmented column of stays through every row operation, so you only ever reduce A itself. Result: here A reduces to the identity (a pivot in all three columns), so are linearly independent . This one test answers several questions at once: independent columns means the vectors form a basis for , that A is invertible, and that every has a unique representation .

This is the written version of the interactive lesson above. See the full Linear Algebra (Matrices) course.