If possible, compute the matrix products in Exercises I through 13, using paper and pencil.

Linear Independence Part 2 Friday, October 14, 2016 3:19 PM n Theorem:Let u ,1… , u me vectorsin R . If m > n then the set of the vectorsis linearly dependent. Proof:0 = ▯ ▯ ▯ + ▯▯ ▯▯ + ⋯+ x ▯▯ ▯▯ ▯ith an augmented matrix of [u ,1… , um, 0] has at most n pivot columns (in echelon form). Since n < m there are m + 1 columns, since n < m there must be at least one column that's not a pivot column (that is also not the last colunm). Since there is at least one solution there is a free variable and thus infinitely many solutions. Thus the set is linearly dependent. Theorem:Let u ,1… , u me a set of vectorsin R . The set is linearly dependent, if and only if one of the vectorsin in the span of the others. Proof (⇒): Assume the set is linearly