Linear Algebra. Chapter 8: Eigenvalues: Further Applications and Computations Section 8.2. Applications to Geometry Proofs of Theorems.


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1 Linear Algebra Chapter 8: Eigenvalues: Further Applications and Computations Section 8.2. Applications to Geometry Proofs of Theorems May 1, 2018 () Linear Algebra May 1, / 8
2 Table of contents 1 Theorem 8.2. Classification of SecondDegree Plane Curves 2 Page 430 Number 16 () Linear Algebra May 1, / 8
3 Theorem 8.2. Classification of SecondDegree Plane Curves Theorem 8.2 Theorem 8.2. Classification of SecondDegree Plane Curves. Every equation of the form ax 2 + bxy + xy 2 + dx + ey + f = 0 for a, b, c not all zero can be reduced to an equation of the form λ 1 t λ 2 t 2 + gt 1 + ht 2 + k = 0 by means of an orthogonal substitution corresponding to a rotation of the plane. The coefficients λ 1 and λ 2 in the second equation are the eigenvalues of the symmetric coefficient matrix of the quadraticform portion of the first equation. The curve describes a (possibly degenerate or empty) ellipse if λ 1 λ 2 > 0 hyperbola if λ 1 λ 2 < 0 parabola if λ 1 λ 2 = 0. () Linear Algebra May 1, / 8
4 Theorem 8.2. Classification of SecondDegree Plane Curves Theorem 8.2 (continued 1) Proof. By Theorem 8.1, Principal Axis Theorem, [ there ] is[ a 2 ] 2 x t1 orthogonal matrix C with det(c) = 1 such that = C and y t 2 ax 2 + bxy + cy 2 = λ 1 t1 2 + λ 2t2 2 where λ 1 and λ 2 are eigenvalues of the symmetric [ coefficient ] matrix of the quadratic form ax 2 + bxy + xy 2. With c11 c C = 12 we have c 21 c 22 [ ] [ ] [ ] [ ] x c11 c = 12 t1 c11 t = 1 + c 12 t 2 y c 21 c 22 t 2 c 21 t 1 + c 22 t 2 and so dx + cy + f = fd(c 11 t 1 + c 12 t 2 ) + e(c 21 t 1 + c 22 t 2 ) + f = (dc 11 + ec 21 )t 1 + (dc 12 + ec 22 )t 2 + f = gt 1 + ht 2 + k. () Linear Algebra May 1, / 8
5 Theorem 8.2. Classification of SecondDegree Plane Curves Theorem 8.2 (continued 1) Proof. By Theorem 8.1, Principal Axis Theorem, [ there ] is[ a 2 ] 2 x t1 orthogonal matrix C with det(c) = 1 such that = C and y t 2 ax 2 + bxy + cy 2 = λ 1 t1 2 + λ 2t2 2 where λ 1 and λ 2 are eigenvalues of the symmetric [ coefficient ] matrix of the quadratic form ax 2 + bxy + xy 2. With c11 c C = 12 we have c 21 c 22 [ ] [ ] [ ] [ ] x c11 c = 12 t1 c11 t = 1 + c 12 t 2 y c 21 c 22 t 2 c 21 t 1 + c 22 t 2 and so dx + cy + f = fd(c 11 t 1 + c 12 t 2 ) + e(c 21 t 1 + c 22 t 2 ) + f = (dc 11 + ec 21 )t 1 + (dc 12 + ec 22 )t 2 + f = gt 1 + ht 2 + k. So the original equation can be reduced to the equation λ 1 t λ 2 t gt 1 + ht 2 + k = 0. () Linear Algebra May 1, / 8
6 Theorem 8.2. Classification of SecondDegree Plane Curves Theorem 8.2 (continued 1) Proof. By Theorem 8.1, Principal Axis Theorem, [ there ] is[ a 2 ] 2 x t1 orthogonal matrix C with det(c) = 1 such that = C and y t 2 ax 2 + bxy + cy 2 = λ 1 t1 2 + λ 2t2 2 where λ 1 and λ 2 are eigenvalues of the symmetric [ coefficient ] matrix of the quadratic form ax 2 + bxy + xy 2. With c11 c C = 12 we have c 21 c 22 [ ] [ ] [ ] [ ] x c11 c = 12 t1 c11 t = 1 + c 12 t 2 y c 21 c 22 t 2 c 21 t 1 + c 22 t 2 and so dx + cy + f = fd(c 11 t 1 + c 12 t 2 ) + e(c 21 t 1 + c 22 t 2 ) + f = (dc 11 + ec 21 )t 1 + (dc 12 + ec 22 )t 2 + f = gt 1 + ht 2 + k. So the original equation can be reduced to the equation λ 1 t λ 2 t gt 1 + ht 2 + k = 0. () Linear Algebra May 1, / 8
7 Theorem 8.2. Classification of SecondDegree Plane Curves Theorem 8.2 (continued 2) Proof (continued). Since det(c) = 1, the transformation x = C t is a rotation (see page 413 of the text). So in the (t 1, t 2 )coordinate system, λ 1 t λ 2t gt 1 + ht 2 + k = 0 is a conic section (possibly degenerate or empty) determined by the coefficients λ 1 and λ 2 as given in the statement of the theorem. () Linear Algebra May 1, / 8
8 Page 430 Number 16 Page 430 Number 16 Page 430 Number 16. Classify the quadric surface with equation 3x 2 + 2y 2 + 6xz + 3z 2 = 1. Solution. To find the symmetric coefficient matrix for the cross terms 3x 2 + 2y 2 + 6xz + 3z 2 = u 11 2 x 2 + u 22 2 y 2 + u 33 2 x 2 + u 12 2 xy + u 21 2 yx + u 13 2 xz + u 31 2 zx + u 23 2 yz + u 32 2 zy, we take u 11 = 6, u 22 = 4, u 33 = 6, u 13 = u 31 = 6, and u 12 = u 21 = u 23 = u 32 = 0. Since a ij = a ji = u ij /2 by Theorem 8.1.A, we have a 11 = 3, a 22 = 2, a 33 = 3, a 13 = a 31 = 3, and a 12 = a 21 = a 23 = a 32 = 0. () Linear Algebra May 1, / 8
9 Page 430 Number 16 Page 430 Number 16 Page 430 Number 16. Classify the quadric surface with equation 3x 2 + 2y 2 + 6xz + 3z 2 = 1. Solution. To find the symmetric coefficient matrix for the cross terms 3x 2 + 2y 2 + 6xz + 3z 2 = u 11 2 x 2 + u 22 2 y 2 + u 33 2 x 2 + u 12 2 xy + u 21 2 yx + u 13 2 xz + u 31 2 zx + u 23 2 yz + u 32 2 zy, we take u 11 = 6, u 22 = 4, u 33 = 6, u 13 = u 31 = 6, and u 12 = u 21 = u 23 = u 32 = 0. Since a ij = a ji = u ij /2 by Theorem 8.1.A, we have a 11 = 3, a 22 = 2, a 33 = 3, a 13 = a 31 = 3, and a 12 = a 21 = a 23 = a 32 = 0. So the symmetric coefficient matrix is a 11 a 12 a A = a 21 a 22 a 23 = a 31 a 32 a () Linear Algebra May 1, / 8
10 Page 430 Number 16 Page 430 Number 16 Page 430 Number 16. Classify the quadric surface with equation 3x 2 + 2y 2 + 6xz + 3z 2 = 1. Solution. To find the symmetric coefficient matrix for the cross terms 3x 2 + 2y 2 + 6xz + 3z 2 = u 11 2 x 2 + u 22 2 y 2 + u 33 2 x 2 + u 12 2 xy + u 21 2 yx + u 13 2 xz + u 31 2 zx + u 23 2 yz + u 32 2 zy, we take u 11 = 6, u 22 = 4, u 33 = 6, u 13 = u 31 = 6, and u 12 = u 21 = u 23 = u 32 = 0. Since a ij = a ji = u ij /2 by Theorem 8.1.A, we have a 11 = 3, a 22 = 2, a 33 = 3, a 13 = a 31 = 3, and a 12 = a 21 = a 23 = a 32 = 0. So the symmetric coefficient matrix is a 11 a 12 a A = a 21 a 22 a 23 = a 31 a 32 a () Linear Algebra May 1, / 8
11 Page 430 Number 16 Page 430 Number 16 (continued 1) Page 430 Number 16. Classify the quadric surface with equation 3x 2 + 2y 2 + 6xz + 3z 2 = 1. Proof (continued). To use Theorem 8.3 (and the not following it) we only need the eigenvalues of A. We have 3 λ 0 3 det(a λi) = 0 2 λ λ = 0+(2 λ) 3 λ λ 0 (2 λ)((3 λ) 2 9) = (2 λ)(λ 2 6λ) = λ(2 λ)(λ 6) and so the eigenvalues are λ 1 = 0, λ 2 = 2, and λ 3 = 6. So, since one eigenvalue is zero and the other two are of the same sign, we know that the surface is either an elliptic paraboloid or an elliptic cylinder. () Linear Algebra May 1, / 8
12 Page 430 Number 16 Page 430 Number 16 (continued 1) Page 430 Number 16. Classify the quadric surface with equation 3x 2 + 2y 2 + 6xz + 3z 2 = 1. Proof (continued). To use Theorem 8.3 (and the not following it) we only need the eigenvalues of A. We have 3 λ 0 3 det(a λi) = 0 2 λ λ = 0+(2 λ) 3 λ λ 0 (2 λ)((3 λ) 2 9) = (2 λ)(λ 2 6λ) = λ(2 λ)(λ 6) and so the eigenvalues are λ 1 = 0, λ 2 = 2, and λ 3 = 6. So, since one eigenvalue is zero and the other two are of the same sign, we know that the surface is either an elliptic paraboloid or an elliptic cylinder. () Linear Algebra May 1, / 8
13 Page 430 Number 16 Page 430 Number 16 (continued 2) Page 430 Number 16. Classify the quadric surface with equation 3x 2 + 2y 2 + 6xz + 3z 2 = 1. Proof (continued). To further narrow the answer, we apply Step 3 of Diagonalizing a Quadratic Form f ( x) from Section 8.1. We have that 3x 2 + 2y 2 + 6xz + 3y 2 = λ 1 t λ 2 t λ 3 t 2 3 2t t 2 3. So in the (t 1, t 2, t 3 )coordinate system, the given equation becomes 2t t2 3 = 1, or t2 2 /3 + t2 3 /1 = 1, or t 2 2 ( 3) 2 + t2 3 (1) 2 = 1. So the surface is in fact an elliptic cylinder (with semimajor axis 3 and semiminor axis 1). () Linear Algebra May 1, / 8
14 Page 430 Number 16 Page 430 Number 16 (continued 2) Page 430 Number 16. Classify the quadric surface with equation 3x 2 + 2y 2 + 6xz + 3z 2 = 1. Proof (continued). To further narrow the answer, we apply Step 3 of Diagonalizing a Quadratic Form f ( x) from Section 8.1. We have that 3x 2 + 2y 2 + 6xz + 3y 2 = λ 1 t λ 2 t λ 3 t 2 3 2t t 2 3. So in the (t 1, t 2, t 3 )coordinate system, the given equation becomes 2t t2 3 = 1, or t2 2 /3 + t2 3 /1 = 1, or t 2 2 ( 3) 2 + t2 3 (1) 2 = 1. So the surface is in fact an elliptic cylinder (with semimajor axis 3 and semiminor axis 1). () Linear Algebra May 1, / 8
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