Homework 7 Solutions

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$Math 33B/1 S00 K. Solutions to Midterm Exam G1, Version B If you ...$

$Math 3C Homework 3 Solutions$

⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞⎫ 7 − (-1) -4 0 8 -4 0 ⎨ 2 ⎬ E-1 = N ⎝ 8 -5 − (-1) 0 ⎠ = N ⎝8 -4 0⎠ = span ⎝4⎠ ⎩ ⎭ 6 -6 3 − (-1) 6 -6 4 3 . The two eigenspaces have a total dimension of 3, so A is diagonalizable. Our diagonalization is therefore ⎛ 1 0 ⎞ ⎛ 2 3 0 ⎞ ⎛ 0 1 0 ⎞-1 2 A = ⎝1 0 4⎠ ⎝0 3 0 ⎠ ⎝1 0 4⎠ =: QDQ 0 1 3 0 0 -1 0 1 3 -1 . ⎛ 1 1 ⎞ 0 f. A = ⎝0 1 2⎠ 0 0 3 Solution: Since A is an upper-triangular matrix, we can read its eigenvalues off the diagonal: λ = 1, 3. Then ⎛ 1 − 1 E1 = N ⎝ 0 0 1 1 − 1 0 ⎞ ⎛ 0 0 2 ⎠ = N ⎝0 3 − 1 0 1 0 0 ⎞ ⎧⎛ 0 ⎨ 2⎠ = span ⎝ ⎩ 2 1 ⎞⎫ ⎬ 0⎠ ⎭ 0 . We needn’t bother to compute E3. The dimension of E1 is 1, although the root λ = 1 has multiplicity 2. Therefore A is not diagonalizable. 3b. Let V = P2(R). Define T : V → V by the mapping T(ax 2 + bx + c) = cx 2 + bx + a. If diagonalizable, find a basis β for V such that [T]β is a diagonal matrix. Solution: If α = {1, x, x2 } is the standard basis on P2(R), then ⎛ 0 A := [T]α = ⎝0 0 1 ⎞ 1 0⎠ . 1 0 0 Computing eigenvalues of A: ⎛ 0 − λ 0 1 ⎞ χA(λ) = det ⎝ 0 1 − λ 0 ⎠ = (0 − λ) 1 0 0 − λ 2 (1 − λ) − (1 − λ) = (1 − λ)(λ 2 − 1) = -(1 − λ) 2 (1 + λ). The roots of this polynomial are λ = ±1. Now, ⎛ ⎞ ⎛ ⎞ ⎧⎛ 0 − 1 0 1 -1 0 1 ⎨ E1 = N ⎝ 0 1 − 1 0 ⎠ = N ⎝ 0 0 0 ⎠ = span ⎝ ⎩ 1 0 0 − 1 1 0 -1 1 ⎞ ⎛ ⎞⎫ 0 ⎬ 0⎠ , ⎝1⎠ ⎭ 1 0 ⎛ ⎞ ⎛ ⎞ ⎧⎛ ⎞⎫ 0 − (-1) 0 1 1 0 1 ⎨ 1 ⎬ E-1 = N ⎝ 0 1 − (-1) 0 ⎠ = N ⎝0 2 0⎠ = span ⎝ 0 ⎠ ⎩ ⎭ 1 0 0 − (-1) 1 0 1 -1 . 2
Thus β = {(1, 0, 1), (0, 1, 0), (1, 0, -1)} is a diagonalizing basis of LA. Noting that [LA]β = [L [T]α ]β = [φα ◦ T ◦ φ -1 α ]β = [T] φ -1 α (β), (1) we know that φ -1 α (β) = {1 + x 2 , x, 1 − x 2 } is a diagonalizing basis of T (equation (1) justifies the obvious final step of converting the vectors of β into their corresponding polynomials). 8. Suppose that A ∈ Mn×nn(F ) has two distinct eigenvalues, λ1 and λ2, and that dim(Eλ1 ) = n − 1. Prove that A is diagonalizable. Solution: Distinct eigenspaces intersect trivially, and any eigenspace has dimension ≥ 1, so dim(Eλ1 + Eλ2 ) = dim(Eλ1 ⊕ Eλ2 ) = dim(Eλ1 ) + dim(Eλ2 ) ≥ (n − 1) + 1 = n. The eigenspaces of A span F n , and so A is diagonalizable. 11. Let A be an n × n matrix that is similar to an upper triangular matrix, and has the distinct eigenvalues λ1, λ2, . . . , λk with corresponding multiplicities m1, m2, . . . , mk. Prove the following statements. Lemma: If A and B are similar matrices, then χA(λ) = χB(λ). Let A = Q -1 BQ for some matrix Q ∈ Mn×n(F). By the multiplicative property of determinants, χA(λ) = det(A − λI) = det(Q -1 BQ − λI) = det(Q -1 (B − λI)Q) = det(B − λI) = χB(λ). Solution: Let M be an upper-triangular matrix such that A = QMQ -1 . It was proved (5.4:9) that the eigenvalues of M coincide with the diagonal elements Mii. Since similar transformations have the same characteristic polynomials (lemma, above), they share eigenvalues λi and multiplicities mi. a. trace(A) = k i=1 miλi Solution: By (2.5:10), b. det(A) = (λ1) m1 (λ2) m2 · · · (λn) mn . trace(A) = trace(M) = n Mii = i=1 k miλi. (2) Solution: The determinant of an upper-triangular matrix is the product of its diagonal entries Mii (determinant property 4). By the multiplicative property of determinants, det(A) = det(QMQ -1 ) = det(M) = i=1 n Mii = i=1 n i=1 λ mi i . (3) Finally, one can show (the proof is a little too complicated to show here, but see pp.370,385 in the text) that every matrix is similar, over some field, to an upper-triangular matrix. The identities (2) and (3) above are therefore universal properties of matrices. 3
Page 1: 5.2 - Diagonalizability Homework 7
Page 5 and 6: Solution: Let β be a basis of V th
Page 7: so ak and bk−1 are orthogonal. Ap

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