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Solutions 10

# Solutions 10 - MASZlZ Linear Algebra I Exercises 10(with...

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Unformatted text preview: MASZlZ Linear Algebra I Exercises 10 (with solutions) © Francis J. Wright, 2006 Assessed submission deadline 5pm on Friday 15th December 2006 in the blue box on the ground ﬂoor. Your submission must have your FULL NAME and STUDENT NUMBER PRINTED CLEARLY at the top and separate sheets must be stapled together. Marks out of 100 are shown in square brackets after each question number. The ﬁeld of scalars is the complex ﬁeld C in all cases. 4 1 —2 1. [25] For the matrixA= —2 1 2 : 0 0 1 (a) construct the characteristic equation of A (by evaluating the determinant in a way that gives the characteristic polynomial in a factorised form if possible); (b) solve this equation to ﬁnd the eigenvalues of A; (c) for each eigenvalue, determine a corresponding eigenvector of A; ((1) write down an invertible matrix P_1 such thatPA P_1 = A is diagonal, write down what A should be, and then check that P A P_1 = A; (e) use A (or equivalently the eigenvalues) to compute the trace and determinant of A. V Solution 4 _)\4 1 —2 (a) The characteristic polynomial of A is —2 1 —7c 2 . Laplace expansion about the 0 01—h bottom row give this in factorised form as ( 1 —7\.) ( (4 —7\.) ( 1 —71.) + 2). Expanding and factorising the second factor leads to the characteristic equation in the form (1 —A) (12—51%) = (1 a) (1,—2) (x—3) =0. (b) Hence the eigenvalues of A are 21. = 1, 2, 3. (c) Solving (A —?»I) x = 0 forx with 2» = 1, 2, 3 gives the eigenvectors (1, —1,1), (—1,2,0), (—1, 1, 0) (up to scalar factors). 1 —1 —1 (d) Using these eigenvectors as the columns gives P_1 = —1 2 1 ,which reducesA to the 1 0 0 1 0 0 diagonal matrix of eigenvaluesA= 0 2 0 . Check: 0 0 3 001 41—2 1—1—1 100 PAP—1:110.—212.—121=020. —2—11 001 100 003 L(e) tr(A)=l+2+3=6anddet(A)=1><ZX3=6. > Maple Solution 1 5 2. [25] Repeat Question 1 withA = 1 5 . V Solution (a) The characteristic equation of A is 9.2—6 1 + 10 = 0 (b) Solving using the quadratic formula, the eigenvalues ofA are 7x. = 3 i i. (c) Substituting each eigenvalue into the equation (A 4» I ) x = 0 and solving the ﬁrst scalar equation for x gives the eigenvectors (5, 2 i i). 5 5 ((1) Using these eigenvectors as the columns gives P_1 = 2 + _ 2 . , which reducesA to the 1 —1 3 +1 0 diagonal matrix of eigenvalues A = 3 . . Check: —1 1 1 . 1 . _1 1o+5121 15 5 5 3+10 PA P = . . _ _ = _ . i—ii ii —1 5 2+1 2—1 0 3—1 10 5 2 (e) tr(A) = (3 +1) + (3—1) =6 anddet(A) = (3 +1) (3—1) =9 +1 =10. But note that eigenvectors are not unique, and a possible alternative pair of eigenvectors that arise naturally from solving the second rather than the ﬁrst scalar equation is (2 \$ 1, 1 ) giving _1 2—i 2+i P = .Check: 1 1 i1 L_i _1 2 2 15 2—12+i 3+1 0 PAP = . . = _ . —ii i+i —1 5 1 1 0 3—1 2 2 Note that (5, 2 i i is a (complex) scalar multiple of (2 \$1, 1) since 2 (Zii) (2\$i, 1) =(5,2ii)- > Maple Solution 3 —2 3 3. [25] Repeat Question 1 withA = 1 0 3 . Hints: To solve the characteristic equation ﬁrst 1 —2 5 show that one of the eigenvalues is 4. It is possible to diagonalise a matrix that has non-distinct eigenvalues if enough eigenvectors can be chosen to be linearly independent. V Solution (a) The characteristic equation of A is 73—8 73 + 20 1—16 =0 which factorizes as (1—4) (Hf: (b) Hence the eigenvalues of A are 11— — 4, 2, 2. (c) Solving (A— M) x= 0 forx with 9t= 4 gives the eigenvector (1, 1,1 )0 and with k= 2 gives the general solution (2y— —,32 y, z), from which we can choose (2,1,0), (—3 , 1) as a basis for the 2- dimensional eigenspace. 1 2 —3 ((1) Using these eigenvectors as the columns gives P_1 = 1 1 0 ,which reducesA to the 1 0 1 4 0 0 diagonal matrix of eigenvaluesA= 0 2 0 . Check: 0 0 2 e —1 g 3 —2 3 1 2 —3 4 0 0 PAP -%2—%103110=020 _i1_i 1—25 101 002 2 2 7(e) tr(A)=4+2+2=8anddet(A)=4x2x2=16. > Maple Solution 2 —l —l 4. [25] The eigenvalues of A = 0 3 l satisfy its characteristic equation, namely 0 l 3 (K—Z ) 2 (k—4) = 0. Verify the Cayley-Hamilton Theorem for this matrix and ﬁnd the lowest degree factor of the characteristic polynomial that vanishes when A is substituted for A. (and constants are multiplied by the identity matrix]3 as necessary). The characteristic polynomial still factorizes when A is substituted for 2» (because all products involved commute) but now no linear factor vanishes even though a product of linear factors vanishes. Verify this and explain it in view of the fact that when A is a ﬁeld element at least one linear factor of the characteristic polynomial must vanish for the characteristic equation to be satisﬁed. VSolution 2 —l —1 0 —1 —l —2 —1 —1 IfA:= O 3 l thenA—2= 0 l 1 andA—4= 0 -1 l 0 l 3 0 l l 0 l —1 Hence neither linear factor of the characteristic equation vanishes for 9» =A. 0 —2 —2 0 0 0 There are two quadratic factors: (A—2)2= 0 2 2 and (A—2) (A—4) = 0 0 0 . 0 2 2 0 0 0 Hence (Qt—2) (7t—4) is the lowest degree factor of the characteristic polynomial that vanishes for 7» =A. Therefore (Qt—2 ) 2 (7L—4) also vanishes for 7L =A, so the Cayley-Hamilton Theorem is satisﬁed. The reason why no linear factor vanishes is that it is possible for the product of two nonzero matrices to be zero (as illustrated here), which cannot happen in a ﬁeld. (The set of all n x n matrices forms a ring with zero divisors, not a ﬁeld.) The rule for ﬁelds that if a product vanishes then at least one factor must vanish does not apply to rings, and hence not to rmatrices. > Maple Solution ...
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Solutions 10 - MASZlZ Linear Algebra I Exercises 10(with...

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