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MIT MATH 52 - SOLUTIONS PROBLEMS 21-35

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MA 52 May 13, 2006Chapters 5,6,7 ReviewSOLUTIONS PROBLEMS 21-35Math 52 Spring 200621. Compute the determinant of the matrixA =0 0 1 00 1 0 01 0 0 00 0 0 13 −2 6 40 1 13 10 0 −2 60 0 0 41 0 0 00 1 0 03 0 1 00 0 0 1.Is A invertible or not? Why?The determinant of the three matrices can b e done individually. For thefirst, it can be reduced to the identity matrix by one row swap, so it hasdeterminant −1. The second is upper triangular and its determinant isthe product of the diagonal entries: (3)(1)(−2)(4) = −24. The third islower triangular and has determinant (1)(1)(1)(1) = 1. Therefore thedeterminant of A is 24.Since the determinant is nonzero, A is invertible.22. Let A =2√2√6√2−2√2√6√20 −2 2√3.(a) Compute AAT.(b) What is A−1?(a) AAT= 16I(b) A−1=116AT1MA 52 May 13, 200623. True or false? The product of any two orthogonal matrices is orthogo-nal.True. An orthogonal matrix is the matrix of an orthogonal transfor-mation. Orthogonal transformations preserve length. The product oftwo orthogonal matrices is the composition of two orthogonal trans-formations, and since each preserves length, so does their composition.Therefore the product of two orthogonal matrices is orthogonal.24. Define each of the following terms and in each case give a 2×2 example:(a) diagonal matrix(b) upper triangular matrix(c) diagonalizable matrix(a) A diagonal matrix is a square matrix whose off-diagonal entriesare all zero. For example,'3 00 −2(.(b) An upper triangular matrix is a square matrix whose entries belowthe diagonal are all zero. For example,'3 70 −2(.(c) A diagonalizable matrix is a square matrix A such that A =SDS−1for some invertible matrix S and diagonal matrix D. Forexample,A ='−2 3−4 5(='1 31 4('1 00 2('4 −3−1 1(Note that one can create such examples by choosing any diagonalD and invertible S and calculating A.25. Let A be an n × n matrix.(a) Suppose !v is an eigenvector of A with eigenvalue λ. Show that !vis also an eigenvector of A2.(b) Prove that if A is diagonalizable, then so is A2.2MA 52 May 13, 2006(a) Suppose !v is an eigenvector of A with eigenvalue λ. Then, A!v =λ!v. So A2!v = AA!v = Aλ!v = λA!v = λ2!v. This shows that !v is aneigenvector of A2with eigenvalue λ2.(b) If A is diagonalizable, then Rnhas a basis of eigenvectors of A.These eigenvectors are all eigenvectors of A2and so form a basisof eigenvectors that will also diagonalize A2.26. Let A and B be two similar matrices.(a) Show that A and B have the same characteristic polynomial.(b) Prove that A and B have the same eigenvalues.(a) Suppose that A and B are two similar matrices. Then A = SBS−1for some invertible S. ThenA −λI = SBS−1− λI= SBS−1− λSIS−1= S(B − λI)S−1by distributivityTherefore A − λI and B − λI are similar. Since similar matriceshave the same determinant, A and B have the same characteristicpolynomial.(b) Since A and B have the same characteristic polynomial, they havethe same eigenvalues.27. Decide if the matrix A =1 0 20 1 30 0 1is diagonalizable. Justify youranswer.The matrix has exactly one eigenvalue with algebraic multiplicity three(since it is upper triangular, its diagonal entries give the eigenvalues).Therefore it will be diagonalizable if and only if the eigenvalue 1 hasgeometric multiplicity 3. But the matrix A−I =0 0 20 0 30 0 0has rank1 and so has nullity 2. So it is not diagonalizable.3MA 52 May 13, 200628. Find an orthogonal basis for the subspace of R4spanned by−1311,6−8−2−4and636−3.Using Gram-Schmidt, one obtains!u1=1√12−1311, !u2=1√12311−1, !u3=1√12−1−13−129. Find the eigenvalues of4 −1 62 1 62 −1 8. Hint: they are integers.The eigenvalues are 2, 2 and 9.30. The matrix A =4 3 3−12 −8 −66 3 1is diagonalizable and has eigenval-ues −2, −2, 1. Find a matrix which diagonalizes A.The eigenspaces areE−2= ker6 3 3−12 −6 −66 3 3= span1−20,10−2E1= ker3 3 3−12 −9 −66 3 0= span−12−1So A = SDS−1where S =1 1 −1−2 0 20 −2 −1and D =−2 0 00 −2 00 0 1.4MA 52 May 13, 200631. Find an orthonormal basis for the subspace of R4spanned by1100,0110and0011.Using Gram-Schmidt, we obtain1√21100,1√6−1120,12√31−113.32. Let V be the vector space consisting of all polynomials p(x) = a0+a1x + a2x2+ a3x3of degree ≤ 3. Let T be the linear transformationT (p(x)) = x2d2pdx2+ p(x).(a) Find the matrix A associated to T for some suitable basis of V .(b) For which real numbers λ does there exist a non-zero solution p(x)to the equationx2d2pdx2+ p(x) = λp(x)?For each such λ find the corresponding p(x).(a) Use the basis B : 1, x, x2, x3. ThenT (a+bx+cx2+dx3) = 2cx2+6dx3+a+bx+cx2+dx3= a+bx+3cx2+7dx3So the transformation has matrix[T ]B=1 0 0 00 1 0 00 0 3 00 0 0 75MA 52 May 13, 2006(b) The question is asking for the eigenvalues of the transformationT . Since the matrix is already diagonal, we see the eigenvaluesare 1, 3 and 7. To find the eigenpolynomials associated to theeigenvalues, we could calculate the eigenspaces directly. However,since this matrix is already diagonal, the basis B is our eigenbasisand so we haveE1= span {1, t}E3= span/t20E7= span/t3033. If A ='1 11 2(and B ='1 22 0(, find all numbers c for which theequation A!x = cB!x has a nonzero solution. For each such c, find thecorresponding !x.We rearrange the equation by multiplying on the left by B−1. We getB−1A!x = c!xso the values of c we seek are the eigenvalues of the transformationB−1A =−14'0 −2−2 1('1 11 2(='121140(This transformation has characteristic polynomial (12− c)(−c) −14=c2−12c −14. This one requires the quadratic formula. The answersare c =1±√−34. Since these are complex numbers, this problem reallyshould not have been included in the review sheet. Nevertheless, thesetup is interesting, and from here on, had there been real eigenvalues,the process would have been routine.34. Let !v and !w be eigenvectors of A


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