By J Whitesitt

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Additional info for Principles of modern algebra

Example text

Then, since C(B) ⊂ C(A, B), C(B) ⊂ C(A). Thus, we have established that C(A) C(A, B) ⇔ C(B) ⊂ C(A). A That C(B) C(A, B) ⇔ C(A) ⊂ C(B), that R(A) R ⇔ R(C) ⊂ C A R(A), and that R(C) R ⇔ R(A) ⊂ R(C) can be established via similar C arguments. D. 2. Let A represent an m × n matrix, B an m × p matrix, and C a q × n matrix. Then, (1) rank(A) ≤ rank(A, B), with equality holding if and only if C(B) ⊂ C(A); (2) rank(B) ≤ rank(A, B), with equality holding if and only if C(A) ⊂ C(B); A , with equality holding if and only if R(C) ⊂ R(A); (3) rank(A) ≤ rank C A (4) rank(C) ≤ rank , with equality holding if and only if R(A) ⊂ R(C).

Xr are linearly dependent. Then, there exist scalars 0 and hence [in y1 , y2 , . . 1)] such that rj 1 yj Cj 0. Thus, C1 , C2 , . . , Cr are linearly dependent. Alternatively, suppose that A1 , A2 , . . , Ak are linearly independent and x1 , x2 , . . , xr are linearly independent. And, let y1 , y2 , . . , yr represent any scalars such that rj 1 yj Cj 0. 1) (and the linear independence of A1 , A2 , . . , Ak ), rj 1 yj xij 0 (for i 1, 2, . . , r) and hence r 0, implying (in light of the linear independence of x1 , x2 , .

4. For any m × n matrix A, m × p matrix B, and n × p matrix L, C(A, B) C(A, B − AL), rank(A, B) rank(A, B − AL) . Similarly, for any m × n matrix A, q × n matrix B, and q × m matrix L, R A B R A , B − LA rank A B rank A B − LA . Proof. 2) that C(A, B − AL) ⊂ C(A, B) and C(A, B) ⊂ C(A, B − AL) and, consequently, that C(A, B) C(A, B − AL) and rank(A, B) rank(A, B − AL). The second part of the lemma can be established in similar fashion. D. 5. (1) If A is an m × n matrix and E an m × q matrix such that C(E) ⊂ C(A) and if B is an m × p matrix and F an m × r matrix such that C(F) ⊂ C(B), then C(E, F) ⊂ C(A, B), rank(E, F) ≤ rank(A, B) .