25 in the new coordinate system y1 y2 it is easy

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Unformatted text preview: length elements of W1 which are perpendicular to each other. Next, we consider the eigenspace W9 corresponding to the eigenvalue λ2 = 9, which consists of the solutions to the linear system 5−9 4 0 b1 4 5−9 0 b2 = 0, 0 0 1−9 b3 or −4b1 4b1 +4b2 −4b2 −8b3 = 0, = 0, = 0. The coefficient matrix of this linear system is −4 4 0 4 −4 0 . 0 0 −8 34 Applying the elementary row operations to this matrix yields −4 4 0 1 −1 0 1 −1 4 −4 0 → 4 −4 0 → 0 0 0 0 −8 0 0 −8 00 1 −1 0 1 −1 0 → 0 0 0 → 0 0 1 . 001 000 0 0 −8 Thus the linear system is equivalent to b1 − b2 b3 0 and we see that We set = 0, = 0, = 0, 1 W9 = span 1 . 0 b3 = 1 √ 2 1 √ 2 . 0 Theory guarantees that the matrix 1 0 √2 − B = 0 √1 2 10 1 √ 2 1 √ 2 , 0 whose columns are the eigenvectors b1 , b1 , and b3 , will satisfy 100 B −1 AB = 0 1 0 . 009 Moreover, since the eigenvectors we have chosen are of unit length and perpendicular to each other, the matrix B will be orthogonal. Exercises: 2.1.1. Find a 2 × 2-matrix B such that B is orthogonal and B −1 AB is diagonal, whe...
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