Note since det a det a t we have the same rules for

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Note: Since det( A ) = det( A T ), we have the same rules for column operations as we are used to for row operations. (That’s because a column operation on A is equivalent to a row operation on A T .) (c) We use row operations to transform the matrix into an upper triangular matrix: A = 1 1 1 1 1 1 4 4 1 - 1 2 - 2 1 - 1 8 - 8 R 2 R 2 - R 1 ,R 3 R 3 - R 1 ,R 4 R 4 - R 1 ----------------------→ 1 1 1 1 0 0 3 3 0 - 2 1 - 3 0 - 2 7 - 9 R 4 R 4 - R 3 ,R 3 R 2 ------------→ 1 1 1 1 0 - 2 1 - 3 0 0 3 3 0 0 6 - 6 R 4 R 4 - 2 R 3 -------→ 1 1 1 1 0 - 2 1 - 3 0 0 3 3 0 0 0 - 12 Since we swap rows once, and the other row operations that we used do not change the value of the determinant, we have: det( A ) = - [1 . ( - 2) . 3 . ( - 12)] = - 72 (d) We use row operations to transform the matrix into an upper triangular matrix: A = 0 0 3 1 0 0 2 2 2 - 1 1 1 1 0 - 1 0 R 3 R 3 - 2 R 4 ,R 1 R 1 - 3 / 2 R 2 -----------------→ 0 0 0 - 2 0 0 2 2 0 - 1 3 1 1 0 - 1 0 R 1 R 4 ,R 2 R 3 ---------→ 1 0 - 1 0 0 - 1 3 1 0 0 2 2 0 0 0 - 2
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Since we swap rows twice, and the other row operations that we used do not change the value of the determinant, we have: det( A ) = - ( - [1 . ( - 1) . 2 . ( - 2)]) = 4 [Note that this determinant is also pleasant to compute by expanding along the second column. Do it!] Problem 6. Let A = - 3 4 1 0 0 2 (a) Find the eigenvalues of A , as well as a basis for the corresponding eigenspaces. (b) Diagonalize A . (That is, write A = PDP - 1 where D is diagonal.) - 1 0 1 . Solution.
Problem 7. Consider the vector space V = { f : R R : f is 7-periodic and f is “nice” } . Here, f “nice” means, for instance, that f should be piecewise continuous or (more generally) that R 7 0 f ( t ) 2 dt should be finite. (a) What is a natural inner product on V ? (b) Consider the 7-periodic function f ( t ) with f ( t ) = 1 , for 0 t < 3 , 2 , for 3 t < 7 .

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