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Differential Equations Solutions 61

# Differential Equations Solutions 61 - results are much...

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71 To prove the second fact, note that trace( CD ) = m k =1 ( n i =1 ( c ki d ik )) , while trace( DC ) = n i =1 ( m k =1 ( d ik c ki )) , which is the same. (b) Note that B QA 2 F = trace(( B QA ) T ( B QA )) = trace( B T B + A T A ) 2 trace( A T Q T B ) , so we can minimize the left-hand side by maximizing trace( A T Q T B ). (c) We compute trace( Q T BA T ) = trace( Q T U Σ V T ) = trace( V T Q T U Σ) = trace( Z Σ) = m i =1 σ i z ii m i =1 σ i , (12.1) where the inequality follows from the fact that elements of an orthogonal matrix lie between 1 and 1. (d) Since Z = V T Q T U , we have Z = I if Q = UV T . CHALLENGE 12.4. The results are shown in Figure 12.1.
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Unformatted text preview: results are much better than those of Challenge 2, with errors at most 10 − 14 and no trouble when the pitch is close to vertical. CHALLENGE 12.5. We compute ± B − QA − te T ± 2 F = m X i =1 n X j =1 ( B − QA ) 2 ij − 2 t i ( B − QA ) ij + nt 2 i , and setting the partial derivative with respect to t i to zero yields t i = 1 n n X j =1 ( B − QA ) ij ....
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