Math 215 HW #9 Solutions

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Doris Alberta Watkins
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1 Math 5 HW #9 Solutions. Problem If A is a 5 by 5 matrix with all a ij, then det A. Volumes or the big formula or pivots should give some upper bound on the determinant. Answer: Let v i be the ith column of A. Then v i = a i + a i + a i + a 4i + a 5i. Since a ij for each i and j, it s also true that each a ij. Hence, the right hand side of the above equation is no bigger than = 5, and so we see that v i 5. This means that each edge of the 5-dimensional box spanned by the columns of A is no longer than 5, meaning that the volume of that box can be no bigger than 5 ) 5 = 5 5/ = 5 5. Since det A is exactly equal to the volume of the box spanned by the columns of A, this reasoning implies that det A Another way to get an upper bound is to use the big formula for the determinant from p.. Notice that, for any 5 by 5 matrix, there are 5! = terms in the sum. When all entries of A are smaller than in absolute value), it must be the case that each of the terms in the sum is smaller than in absolute value). Therefore, the whole sum which is just the determinant), must be no bigger than in absolute value.. Problem Find A from the cofactor formula C T / det A. Use symmetry in part b): a) A = 4. b) A = a) We need to determine the various cofactors of A to find the cofactor matrix C: C = ) + det A = 4 = C = ) + det A = = C = ) + det A = 4 = C = ) + det A = 4 =.

2 Therefore, C = ) + det A = = C = ) + det A = 4 = 4 C = ) + det A = = C = ) + det A = = C = ) + det A = = C = C C C C C C C C C = 4 Also, since we ve already computed the relevant cofactors, it s easy to find det A: det A = a C + a C + a C = + + =.. Therefore, A = det A CT = 4 = / / 4/ b) Again, we need to determine the various cofactors of A to find the matrix C, but now we can use the fact that A is symmetric, which implies that A ij = A T ji and so C ij = ) i+j det A ij = ) i+j det A T ji = ) j+i det A ji = C ji. Thus, we only need to compute the C ij for which i j: C = ) + det A = = C = ) + det A = = C = ) + det A = = C = ) + det A = = 4 C = ) + det A = = C = ) + det A = =

3 Therefore, Also, so we have that C = C C C C C C C C C = 4 det A = a C + a C + a C = + ) + = 4, A = det A CT = 4 4 =. /4 / /4 / / /4 / /4. Problem If all entries of A are integers, and det A = or, prove that all entries of A are integers. Give a by example. Proof. If all entries of A are integers, then every cofactor of A is an integers since each cofactor is, ultimately, a sum of products of integers). Therefore, the entries in the cofactor matrix C are all integers, which means that the entries in C T are also integers. Moreover, since det A = ±, we see that A = det A CT = ±C T, so all of the entries of A are integers. Example: Consider the by matrix A = [ 5 9 Then det A = 5 9 = 5 4 =, and so A = det A CT = C C = = det A C C 9 5 which is certainly a matrix with integer entries. ]. [ Problem A box has edges from,, ) to,, ),,, ), and,, ). Find its volume and also find the area of each parallelogram face. Answer: If we form the matrix A = then the columns of A form the edges of the box, and so the determinant of A will give the volume of the box: ], det A = a C + a C + a C = + = 8 + ) =.

4 Therefore, the volume of the box is. Notice that, if we can find a vector v of length which is perpendicular to the side spanned by the first two columns of A call them a and a ), then the volume of the box spanned by a, a, and v will be the same as the area of the parallelogram spanned by a and a. To find such a v, note that the nullspace of the matrix [ ] is orthogonal to its row space, which is just the subspace spanned by a and a. elimination on the above matrix yields /4, 8/ / Doing so the homogeneous solutions i.e. vectors in the nullspace) are of the form /4 x /4 for some real number x. Such vectors have length [ x + 4) ) ] + 4 = x = x 8 = x. Thus, letting x =, we get which has length and is perpendicular to a and a. Therefore, the area of the parallelogram spanned by a and a is equal to det = + = = To find the area of the parallelograms spanned by a and a and by a and a we can use the exact same reasoning with a rearrangement of the components since a, a, and a are the same up to rearrangement of their components), so these other parallelograms also have area. 4

5 5. Problem If the columns of a 4 by 4 matrix have lengths L, L, L, L 4, what is the largest possible value for the determinant based on volume)? If all entries are or, what are those lengths and the maximum determinant? Answer: If the four columns have lengths L, L, L, L 4, then the four-dimensional box spanned by the columns of the matrix has edges of length L, L, L, L 4. Thus, the volume of this box is no larger than the product L L L L 4. Since the absolute value of the) determinant of a matrix is equal to the volume of the box spanned by the columns, this implies that det A L L L L 4. If all entries of the matrix are or, then each L i is equal to ±) + ±) + ±) + ±) = 4 =, and so det A 4 =. In fact, the following is such a matrix with determinant exactly equal to :.. Problem Show that the determinant equals the product of the eigenvalues by imagining that the characteristic polynomial is factored into and making a clever choice of λ. deta λi) = λ λ)λ λ) λ n λ), Proof. Suppose that λ,..., λ n are the eigenvalues of the matrix A. Then the λ i are the roots of the polynomial deta λi), meaning that this polynomial factors as deta λi) = λ λ)λ λ) λ n λ). Letting λ which we re thinking of purely as an abstract variable) equal zero and simplifying both sides of the above equation, we see that det A = λ λ λ n, so the determinant of A is equal to the product of the eigenvalues of A. 5

6 7. Problem Find the rank and all four eigenvalues for both the matrix of ones and the checkerboard matrix: A = and C =. Which eigenvectors correspond to nonzero eigenvalues? Answer: Since all columns of A are equal, the rank of A is equal to ; this implies that is an eigenvalue of A, with eigenvectors given by the elements of the nullspace of A. Since elements of the nullspace of A are of the form we see that are linearly independent eigenvectors corresponding to the eigenvalue. x and + x + x 4 To find the other eigenvalue of A since A has rank, there can be at most one), we solve = deta λi) λ = λ λ λ λ = λ) λ λ λ λ λ + λ λ λ = λ) λ) λ λ λ + λ λ ) λ λ + λ + ) λ λ) λ + λ ) λ) λ + )

7 Then, after simplifying, we see that = deta λi) = λ) 4 λ) + 8 λ) = λ 4 4λ = λ λ 4). Therefore, the nonzero eigenvalue of A is 4; the corresponding eigenvector is in the nullspace of A 4I =. This matrix reduces to 8/ 8/ so the nullspace is the line containing the vector associated to the eigenvalue 4. meaning that this is the eigenvector Turning to the checkerboard matrix C, notice that the first two columns are linearly independent, but that the third and fourth columns are repeats of the first two columns, so the matrix must have rank. Therefore, one eigenvalue is ; the corresponding eigenvectors will be the elements of the nullspace of C. Since C reduces to we see that forms a basis for the nullspace of C, meaning that these two vectors are eigenvectors corresponding to the eigenvalue. 7

8 The other two eigenvalues will come from solving = detc λi) λ = λ λ λ λ = λ λ λ λ λ + λ λ = λ λ λ ) λ λ + λ λ λ + λ ) + λ λ + = λ 4 4λ = λ λ 4) = λ λ )λ + ), ) + so the nonzero eigenvalues of C are and. eigenvector will be any vector in the nullspace for You can easily check that C I = is such an eigenvector. For the eigenvalue, the corresponding Likewise, the eigenvector corresponding to the eigenvalue will be any vector in the nullspace for C )I =. You can check that is such an eigenvector. 8. Problem What do you do to A x = λ x in order to prove the following? a) λ is an eigenvalue of A.. 8

9 Proof. Since we know that A x = λ x, we can just multiply both sides of this equation by A: A x = Aλ x) = λa x = λλ x) = λ x where in the second-to-last equality we used the fact that A x = λ x. Thus, we see that x is an eigenvector for the matrix A with corresponding eigenvalue λ, so indeed λ is an eigenvalue for A. b) λ is an eigenvalue of A. Proof. Since we know A x = λ x we can assuming A is invertible) multiply both sides of this equation by A, yielding x = A λ x) = λa x. Since A is invertible, λ, so we can multiply both sides by λ = λ and see that λ x = A x. But this just means that λ is an eigenvalue of A with corresponding eigenvector x. c) λ + is an eigenvalue of A + I. Proof. We know that A x = λ x and of course I x = x, so we have that A + I) x = A x + I x = λ x + x = λ + ) x. Therefore, λ + is an eigenvalue for A + I with corresponding eigenvector x. 9. Problem 5... Solve detq λi) = by the quadratic formula to reach λ = cos θ ± i sin θ: cos θ sin θ Q = rotates the xy-plane by the angle θ. sin θ cos θ Find the eigenvectors of Q by solving Q λi) x =. Use i =. Answer: We have that detq λi) = cos θ λ sin θ sin θ cos θ λ = cos θ λ)cos θ λ) sin θ) sin θ) = cos θ λ cos θ + λ + sin θ = λ λ cos θ + sin θ + cos θ = λ λ cos θ +. Therefore, λ = cos θ ± 4 cos θ 4 = cos θ ± sin θ = cos θ ± i sin θ, 9

10 so indeed the two complex numbers cos θ ± i sin θ are the two eigenvalues of Q. Now, the eigenvector corresponding to the eigenvalue cos θ+i sin θ will be in the nullspace of the matrix i sin θ sin θ Q cos θ + i sin θ)i =. sin θ i sin θ To get this matrix into reduced echelon form, subtract i times row from row to get i sin θ sin θ. Thus, the nullspace is the line spanned by the vector i, which is thus the eigenvector corresponding to the eigenvalue cos θ + i sin θ. Similarly, the eigenvector corresponding to the eigenvalue cos θ i sin θ will be in the nullspace of the matrix i sin θ sin θ Q cos θ i sin θ)i =. sin θ i sin θ Adding i times row to row yields [ i sin θ sin θ Thus, the nullspace is the line spanned by the vector i, which is thus the eigenvector corresponding to the eigenvalue cos θ i sin θ.. Problem 5... Choose the second row of A = so that A has eigenvalues 4 and 7. Answer: If A has eigenvalues 4 and 7, then it must be the case that deta λi) = λ 4)λ 7) = λ λ + 8. On the other hand, we know that, if A =, then a a deta λi) = λ a λ = λa λ) a = λ a λ a. a Therefore, we should pick a = and a = 8, meaning that A =. 8 ].

Conceptual Questions for Review

Conceptual Questions for Review Chapter 1 1.1 Which vectors are linear combinations of v = (3, 1) and w = (4, 3)? 1.2 Compare the dot product of v = (3, 1) and w = (4, 3) to the product of their lengths.