Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# C false they have d 1 18 ab d b 1 bab so ab is similar

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Unformatted text preview: indeﬁnite ( < 0 and  > 0) because the determinant ac b 2 is negative. If c > 9 the graph of z is a bowl, if c < 9 the graph has a saddle point. When c D 9 the graph of z D .2x C 3y/2 is a “trough” staying at zero along the line 2x C 3y D 0. Orthogonal matrices, exponentials e At , matrices with det D 1 are groups. Examples of subgroups are orthogonal matrices with det D 1, exponentials e An for integer n. Another subgroup: lower triangular elimination matrices E with diagonal 1’s. A product AB of symmetric positive deﬁnite matrices comes into many applications. The “generalized” eigenvalue problem K x D M x has AB D M 1 K . (often we use eig.K; M / without actually inverting M .) All eigenvalues  are positive: AB x D x gives .B x /T AB x D .B x /T x: Then  D x T B T AB x =x T B x > 0: p p 3; 2 1; 2; 2 C 1; 2 C 3. The 2 cos k D 2 6 product of those eigenvalues is 6 D det K . 35 Put parentheses in x T AT CAx D .Ax /T C.Ax /. Since C is assumed positive deﬁnite, this energy can drop to zero only when Ax D 0. Sine A is assumed to have independent columns, Ax D 0 only happens when x D 0. Thus AT CA has positive energy and is positive deﬁnite. 34 The ﬁve eigenvalues of K are 2 My textbooks Computational Science and Engineering and Introduction to Applied Mathematics start with many examples of AT CA in a wide range of applications. I believe this is a unifying concept from linear algebra. Problem Set 6.6, page 360 so M D F G 1 . C similar to A and B ) A similar to B .     10 30 01 1 2 AD is similar to B D D M AM with M D . 03 01 10 1 B D GC G  1  D GF 1 AF G 1 Solutions to Exercises 72  1 3 BD 0  1 BD 1  4 BD 2   1   0 10 10 10 D DM 0 11 10 11   1   1 1 0 11 10 D ; 1 0 1 11 0 1   1   3 01 12 01 D . 1 10 34 10 4 A has no repeated  so it can be diagonalized: S  1 5 0  1 0    1 00 10 0 , , , 0 11 10 0   0 0 is by itself and also 1 1 1 AM ; 1 AS D ƒ makes A similar to ƒ. 1 are similar (they all have eigenvalues 1 and 0). 1  1 is by itself with eigenvalues 1 and 1. 0  6 Eight families of similar matrices: six matrices have  D 0, 1 (one family); three matrices have  D 1, 1 and three have  D 0, 0 (twop families each!); one has  D 1 1, 1; one has  D 2, 0; two matrices have  D 2 .1 ˙ 5/ (they are in one family). 1 AM /.M 1 x / D M 1 .Ax / D M 1 0 D 0 (b) The nullspaces of A and of M 1 AM have the same dimension. Differentvectors and different bases.    Same ƒ 01 0 2 have the same line of eigenvectors But A D and B D Same S 00 0 0 and the same eigenvalues  D 0; 0.           12 13 1k 10 11 A2 D , A3 D , every Ak D . A0 D and A 1 D . 01 01 01 01 01 2  k  1  c 2c c kc k 1 c c2 2 k 0 1 JD and J D ; J D I and J D . 0 c2 0 c1 0 ck      du dv 5 v.0/ 1 u.0/ D D . The equation D u has D v C w and 2 w.0/ 0 dt dt dw D w . Then w.t / D 2e t and v.t / must include 2t e t (this comes from the dt repeated ). To match v.0/ D 5, the solution is v.t / D 2t e t...
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## This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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