By Amodei L., Dedieu J.-P.

ISBN-10: 2100520857

ISBN-13: 9782100520855

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These spaces are orthogonal and v T v = 0. 38 30 When AB = 0, the column space of B is contained in the nullspace of A. So rank(B) ≤ 4 − rank (A) = (dimension of the nullspace A). 31 null(N ) produces a basis for the row space of A (perpendicular to N (A)). 2, page 181 1 (a) a T b/a T a = 5/3; p = (5/3, 5/3, 5/3); e = (−2/3, 1/3, 1/3) (b) a T b/a T a = −1; p = (1, 3, 1); e = (0, 0, 0). 2 (a) p = (cos θ, 0) (b) p = (0, 0) since a T b = 0. 1 1 1 5 1 3 1 1 1 1 1 2 3 P1 = 1 1 1 and P1 b = 5 and P1 = P1 .

8 QT Q = I ⇒ |Q|2 = 1 ⇒ |Q| = ±1; Qn stays orthogonal so can’t blow up. Same for Q−1 . 9 det A = 1, det B = 2, det C = 0. 10 If the entries in every row add to zero, then (1, 1, . . , 1) is in the nullspace: singular A has det = 0. ) If every row adds to one, then rows of A − I add to zero (not necessarily det A = 1). 11 CD = −DC ⇒ |CD| = (−1)n |DC| and not −|DC|. If n is even we can have |CD| = 0. 44 12 det(A−1 ) = det d ad−bc −b ad−bc −c ad−bc a ad−bc = ad − bc 1 = . (ad − bc)2 ad − bc 13 Pivots 1, 1, 1 give det = 1; pivots 1, −2, −3/2 give det = 3.

They are a basis for that space. If they are the columns of A then m is not less than n (m ≥ n). 17 These bases are not unique! (a) (1, 1, 1, 1) (c) (1, −1, −1, 0), (1, −1, 0, −1) (b) (1, −1, 0, 0), (1, 0, −1, 0), (1, 0, 0, −1) (d) (1, 0)(0, 1); (−1, 0, 1, 0, 0), (0, −1, 0, 1, 0), (−1, 0, 0, 0, 1). 18 Any bases for R2 ; (row 1 and row 2) or (row 1 and row 1 + row 2). 19 (a) The 6 vectors might not span R4 (b) The 6 vectors are not independent (c) Any four might be a basis. 20 Independent columns ⇒ rank n.

### Analyse numerique matricielle by Amodei L., Dedieu J.-P.

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