Spring 2005 - Nagy's Class - Quiz 4

Spring 2005 - Nagy's Class - Quiz 4 - 2, A does not have a...

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Math 20F Quiz 4 (version 2) May 27, 2005 1. (5.2.18) Find h in the matrix A below such that the eigenspace for λ = 7 is two-dimensional. A = 7 - 1 3 - 6 0 1 h 0 0 0 7 1 0 0 0 3 The eigenspace of A corresponding to λ = 7 is equal to Nul( A - 7 I ). Thus, we must Fnd the value of h for which dimension of Nul( A - 7 I ) is 2, which is the value of h for which the number of free variables in the homegeneous system ( A - 7 I ) x = 0 is 2. A - 7 I = 0 - 1 3 - 6 0 - 6 h 0 0 0 0 1 0 0 0 - 4 , and the reduced echelon form of A - 7 I is 0 1 - 3 0 0 0 18 - h 0 0 0 0 1 0 0 0 0 . Thus, the homogeneous system ( A - 7 I ) x = 0 has 2 free variables and the dimension of the eigenspace of A corresponding to λ = 7 is 2 when h = 18. 2. (5.3.24) A is a 3 × 3 matrix with two eigenvalues. Each eigenspace is one-dimensional. Is A diagonalizable? Why (or why not)? Since the dimensions of the eigenspaces of A add up to only
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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 = 5-6 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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