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51. The eigenspace has dimension 1. Solve (A-4I)v=0; the only solutions are (x 0 0)^T.
52. The eigenspace is all of R3, and so has dimension 3.
53. The top row of A^10 is (-1022 -1023) and the bottom row is (2046 2047).
54. Write A as XDXinv and substitute into the equation given for B: B = SinvXDXinvS = (SinvX)D(SinvX)inv so B is diagonalizable with the same D as A, but with SinvX doing the job of X.
55. Yes, it is a linear transformation.
56. S-perp should be two dimensional, so we should look for two basis vectors. They both must satisfy +1x-1y+3z=0 to be orthogonal to the vector given. Two such vectors are (x y z) = (1 1 0) and (-3 0 1), and they will do perfectly well as a basis for S-perp. (Many other pairs of vectors will also suffice---as long as they're orthogonal to (1 -1 3) and are lin. ind. of each other.)
57a. No.
57b. Vague question, I guess. IF THE VECTORS FORM AN ORTHONORMAL BASIS then Parseval's theorem applies and the length of x is sqrt(3sq + 5sq + (-2)sq) = 3.
58. The first vector is the vector projection of (-1 3 -1) onto (1 -1 3), which is (1 -1 3) times the constant (-7/sqrt(11)), I think. The other vector is whatever it takes to add to (-1 3 -1); check that the second vector is really orthogonal to (1 -1 3).
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