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Unformatted text preview: 2, A does not have a set of 3 linearly independent eigenvectors; thus, A is not diagonalizable. 3. (6.1.26) Let u = 56 7 , and let W be the set of all x in 3 such that u x = 0. What theorem (in Chapter 4) can be used to show that W is a subspace of 3 ? Describe W in geometric language. u x = u T x ; thus, W = Nul ( u T ) . By the theorem (Theorem 2 in Chapter 4) stating that the null space of an m n matrix A is a subspace of n , W is a subspace of 3 since it is the null space of the 1 3 matrix u T . Geometrically, W is the plane in 3 orthogonal to the vector u ....
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 Spring '03
 BUSS
 Math, Linear Algebra, Algebra

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