MT_1-7

# MT_1-7 - Tutorial MT: Matrix Theory 5. Solutions to...

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Unformatted text preview: Tutorial MT: Matrix Theory 5. Solutions to Exercises Exercise 1. 1. From the definition of the scalar product, Eqn. 16, using the summation convention, (A; B) = (abi ) = aib = b ai = (B; A). i i i P 2. (A; A) = a ai = n jaij2 0, since it is the sum of squares. The equality holds only if i i=1 ai = 0 for all i: 3. (A; B) = (ai) b i = abi = (A; B). i Exercise 2. We have (MB)i = Mij bj ; so that (painfully step-by-step) (A; MB) = (MB; A ) = ((MB) ai ) i = (Mij bj ) a i ~ i = Mj iabj ~ = Mji ai b j y = Mji ai b j = MyA b j = My A; B : j Exercise 3. Consider the scalar product of two transformed vectors, UA and UB. From Eqn. 20, (UA; UB) = UyUA; B = U1UA; B = (IA; B) = (A; B) : ~ Uij Ujk = ik = UjiUj k; which is the scalar product of vectors ui = fUji g and uk = fUjk g, i.e., (ui; uk) = ik : These vectors are the columns of the matrix U: Exercise 5. One can, by inspection, write down a vector that is orthogonal to v1 and normalized. For example, take 0 1 0 1 1 1 1 @ A 1 @ v1 = p i ; v2 = p 0 A; 3 2 1 1 13 Exercise 4. Since Uy U =I, we have Tutorial MT: Matrix Theory 1 (v1 ; v2 ) = p6 (1 + 0 1) = 0: Write v3 = a1^1 + a2^2 + a3^3 where ^i is the ith home basis x x x x vector, e.g., 0 1 1 x ^1 = @ 0 A ; 0 etc., and require (v1; v3) = (v2; v3) = 0: Thus, 1 p (a1 ia2 + a3) = 0; 3 1 p (a1 a3) = 0 =) a1 = a3: 2 p Thus, a2 = 2ia1 ; so we can take a1 = 1= 6 and have 0 1 1 1 v3 = p @ 2i A ; 6 1 which we could as easily have also written by inspection. W now form the matrix e V = v1 v2 v3 0 1 1 1 1 and compute the products 0 1 and B Vy V = @ 0 p 3 1 p 2 1 p 6 B = @ p 3 i p 3 1 p 3 p 2 0 1 p2 p 6 2i p6 1 p 6 C A; i p3 0 2i p 6 1 p 3 1 p2 1 p 6 10 CB A@ 10 CB A@ 1 p 3 i p 3 1 p 3 1 p 2 0 1 p2 1 p 6 2i p6 1 p 6 1 1 1 0 0 C @ 0 1 0 A A= 0 0 1 1 1 1 0 0 C @0 1 0A : A= 0 0 1 0 0 Exercise 6. Setting M = 0 in each of Eqns. 23, 24 and 25 results in every term proportional to a power of M becoming the null matrix. Exercise 7. Multiplying each side of the equation by (IR)1, assuming that it exists, we have (IR)1 (IR) M = (IR)1 S B VVy = @ 1 p 3 i p 3 1 p 3 1 p 2 0 1 p2 1 p 6 2i p6 1 p 6 1 p 3 1 p 2 1 p 6 i p3 0 2i p 6 1 p 3 1 p2 1 p 6 14 ...
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## This note was uploaded on 07/13/2008 for the course PHY 201 taught by Professor Covatto during the Spring '08 term at ASU.

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