11 to see that 1 0 3 3 2 c1 0 1 4 c2 1 1

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Unformatted text preview: · · , 44 b1 · bn = 0, b2 · bn = 0, · bn · bn = 1. (2.9) From the discussion in Section 2.1, we recognize that the term orthonormal basis is just another name for a collection of n vectors which form the columns of an orthogonal n × n matrix. It is relatively easy to express an arbitrary vector f ∈ R n in terms of an orthonormal basis b1 , b2 , . . . , bn : to find constants c1 , c2 , . . . , cn so that f = c1 b1 + c2 b2 + · · · + cn bn , (2.10) we simply dot both sides with the vector bi and use (2.9) to conclude that ci = bi · f . In other words, if b1 , b2 , . . . , bn is an orthonormal basis for R n , then f ∈ Rn ⇒ f = (f · b1 )b1 + · · · + (f · bn )bn , (2.11) a generalization of the formula we gave for expressing a vector in R 3 in terms of the standard basis {i, j, k}. This formula can be helpful in solving the initial value problem dx = Ax, dt x(0) = f , (2.12) in the case where A is a symmetric n × n matrix and f is a constant vector. Since A is symmetric,...
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