UCSD ECE 269 Handout # 5

(c) Columns of A are independent. (d) A is tall (i.e., n ≤ m) and full-rank (i.e., rank(A) = min(m,n) = n). Solution: We will show the chain of equivalences (a) =⇒ (b) =⇒ (c) =⇒ (d) =⇒ (a). (a) =⇒ (b): By the rank–nullity theorem, we have dim(N (A)) + rank(A) = n, which implies rank(A) = n (since dim(N (A)) = 0). Since rank(A) = rank(A ), we then have rank(A ) = n. Since rank is equivalent to the dimension of the column space, the dimension of the column space of A is n. Because each column vector in A is of length n, this means that R(AT ) = F. (b) =⇒ (c): Since A is onto, rank(A ) = dim(R(AT )) = n. Because rank(A) = rank(A ) = n, the dim(R(A)) = n. Note now that A has n column vectors and for them to span a space of dimension n, all of these column vectors have to be independent. (c) =⇒ (d): If the columns of A are independent, since each column vector is of length m, there cannot be more than m of them (since more than m vectors of length m necessarily need to be dependent). Thus n ≤ m. Since n independent vectors span a space of dimension n, we know that dim(R(A)) = n =⇒ rank(A) = n = min(m,n). (d) =⇒ (a): By the rank–nullity theorem, rank(A) + dim(N (A)) = n. Since rank(A) = n, we have dim(N (A)) = 0, which implies that N (A) = {0}.