Algebra Workshops 10 and 11

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1 Algebra Workshops 1 and 11 Suggestion: For Workshop 1 please do questions 2,3 and 14. For the other questions, it s best to wait till the material is covered in lectures. Bilinear and Quadratic Forms on Real Vector Spaces 1. (a In lectures an SBF was defined by its properties of (i symmetry, (ii linearity in the first argument. Prove that an SBF is linear in the second entry. (b If B is a symmetric n n matrix, prove that b(x, y = x T By defines a SBF on R n, where x, y are column vectors as usual. (c Let b be a SBF on a vector space V. Show that N b := {x V b(x, v = for all v V } is a subspace of V. (a b(u, λ 1 v 1 + λ 2 v 2 = b(λ 1 v 1 + λ 2 v 2, u (symmetry = λ 1 b(v 1, u + λ 2 b(v 2, u (linearity in the first argument = λ 1 b(u, v 1 + λ 2 b(u, v 2 (symmetry. (b i. (symmetry x T By is a product of 1 n, n n and n 1 matrices, i.e. a 1 1 matrix. Hence it equals its own transpose, i.e. x T By = (x T By T = y T B T (x T T = y T Bx. (Here we have used the rule for transposing a product, and the fact that B is symmetric. ii. (linearity Obvious, really: (λ 1 x 1 + λ 2 x 2 T By = ( (λ 1 x 1 T + (λ 2 x 2 T By = ( λ 1 x T 1 + λ 2x T 2 By = λ 1 x T 1 By + λ 2 x T 2 By. For an n n real symmetric matrix B, its type is (p, q, where B has p positive eigenvalues and q negative eigenvalues. Its signature is p q. These definitions carry over to an SBF b having a matrix B with respect to some basis of V. Also, the rank of b is the rank of B. ( (a Consider the SBF b on R 2 with matrix B =. Write down the corresponding 3 1 quadratic form in terms of standard coordinates x 1, x 2 on R 2. Find a vector v in R 2 such that b(v, v =. Give a sketch showing the regions in the plane where b(x, x is positive, negative and zero. Find the eigenvalues of the matrix, and hence or otherwise determine the type, rank and signature of b. (b Consider the SBF on R 3 with matrix B = Write down the corresponding quadratic form in terms of standard coordinates x 1, x 2, x 3 on R 3. Sketch the regions in R 3 where b(x, x is positive, negative and zero. Find the eigenvalues of the matrix, and hence or otherwise determine the type, rank and signature of b.

2 [Ans: (a 2x x 1x 2 +x 2 2, ( 3+ 7, 2,, (3± 37/2, 1 b < b > 1 1 x b > b < (1,1,2, (b x x 2 2 x 2 3,, {1, 1, 1}, (2,1, 3, 1] 1 b > b > ( ( x1 2x1 + 3x (a The quadratic form is (x 1, x 2 B = (x x 1, x 2 2 = 2x 2 2 3x 1 + x 1+6x 1 x 2 +x To find v we set this to zero: 2x x 1x 2 + x 2 2 = 2(x 1/x (x 1 /x = if x 2. Hence x 1 /x 2 = 3/2 ± 7/2. Taking the plus sign and x 2 = 2 (for example leads to the stated result. There are two lines on which b = (obtained by taking the two signs, and these are sketched in the Maple plot. To find which regions are positive and which negative, just try some simple points. For instance b(1, 1 = 9 >. Eigenvalues are given by (2 λ(1 λ 9 =. Hence stated results. Eigenvalues are of opposite sign (one each, hence p = 1 and q = 1. Rank and signature are p+q and p q respectively. (b Quadratic form obvious. b > iff x 2 3 < x2 1 + x2 2. In cylindrical polar coordinates r, θ, z this is z < r, which is the equatorial zone between the two cones in the figure. b = on their surface, and b < in the polar regions. Eigenvalues are obvious, two positive and one negative. Hence results. 3. (a Let V be the vector space of 2 2 real matrices with real entries and trace zero. What is the dimension of V? (Read (a right through before deciding whether to give up. Consider the SBF (known, by the way, as the trace( form b(x, ( Y = Trace(XY ( on 1 1 V. Find the matrix of b with respect to the basis,, 1 1 of V. Find the eigenvalues of the matrix, and hence or otherwise determine the type, rank and signature of b. (b Let P 2 denote the real vector space of polynomials of degree 2 in a variable x. What is the dimension of P 2? Show that b(p(x, q(x := 1 p (xq (xdx defines an SBF on P 2. (The dashes indicate derivatives. Find a nonzero element of N b (see question 1c and hence deduce that b is degenerate. Find the matrix of b with respect to the basis {x 2, x, 1} of P 2 and hence or otherwise find the type, rank and signature of b. 1 4/3 1 [Ans: (a 3, 1, { 1, 1, 2}, (2,1, 3, 1; (b 3, 1, 1 1, {(7 ± 2 37/6, }, (2,, 2, 2]

3 ( a b (a A general such matrix is A = c a ( 1 = a 1 ( 1 + b ( + c 1 which shows that the three matrices given later are indeed a spanning set; they are also obviously linearly independent, since their general linear combination A equals zero iff a = b = c =. Thus they are a basis, and dimv = 3. Calling the three matrices (as ordered in the question X 1, X 2, X 3, the entries of the matrix B are B ij = b(x i, X j. Thus B 11 = Tr(X1 2 = Tr =. Similar, not always quite so trivial, calculations lead to the stated matrix. The eigenvalues are given by (( λ 2 1(2 λ, hence results. Two are positive, one negative and so the type, rank and signature are as stated. (b P 2 (sometimes called R 2 (x in class has the obvious basis given, hence the dimension. b is obviously symmetric, and linear in the first argument (as both differentiation and integration are linear. Clearly b(1, p = for all p P 2. Hence there is clearly at least a one-dimensional subspace (Sp(1 on which b(p(x, p(x =. Hence the rank of b must be less than 3, i.e. the form is degenerate. The matrix has 11-component B 11 = 1 (2x2 dx = 4/3. Similarly the other entries are easily found, as given in the answer. The eigenvalues are given by ((4/3 λ(1 λ 1( λ =, which leads to the stated results. p = 2, q =, and the other results follow immediately. I p 4. (a Suppose that the matrix of b in a basis is in the standard form I q. Identify a p-dimensional subspace on which b is positive definite. Identify also an (n p-dimensional subspace on which b is negative semi-definite (meaning that b(v, v for all v in the subspace. By considering intersections, deduce that there is no subspace dimension larger than p on which b is positive definite. (b True or false: an SBF is non-degenerate if and only if its matrix does not have zero as an eigenvalue. (c What are the possible types of an SBF on R 3 if there exists a 2-dimensional subspace on which it is negative definite? (d An SBF on R n has type (p, q. What is the largest possible dimension for a subspace V such that b(v, v < for all nonzero v V? (e Same as (d, but now b(v, v for all nonzero v V. (f True or false: If an SBF is positive definite on subspaces U, U V then it is positive definite on their sum U + U. Explain your answer. (Hint: take a look at Q.2. [Ans: (a Sp(v 1,..., v p if p >, Sp(v p+1,...,v n if p < n; (b T; (c (,2, (,3, (1,2; (d q; (e n p; (f F] (a The associated quadratic form is p 1 x2 i p+q p+1 x2 i, and so the subspaces stated are obvious. Let U be a subspace on which b is positive definite, and W the stated subspace on which it is negative semi-definite. On non-zero vectors v U W we must have both b(v, v > and b(v, v. Hence dim(u W =. Also U +W V (the vector space on which b is defined and so dim(u +W n. Using the general result that dim(u + dim(w = dim(u + W + dim(u W we deduce that dim(u dim(v + dim(u W dim(w = n + (n p = p.,

4 (b It is non-degenerate iff, when put in standard form, p + q = n. If this is true, all eigenvalues are non-zero, and conversely. (c Here we know only that q 2, while p + q 3. This leads to the three possibilities listed. All are realisable; for instance the forms x 2 1 x2 2, x2 1 x2 2 x2 3, x2 1 x2 2 +x2 3. (d See the first part of this question (swapping positive and negative. (e Another application of the same idea. There is a subspace of dimension p on which b is positive definite. (f Look at part (a of the question indicated. Choose any two 1-dimensional subspaces U, V lying in the region where b >. Then U +V = R 2, but b is not positive definite on the whole of R (a Find a 2 2 orthogonal matrix P such that P T SP is diagonal, where S = ( (b Find also a non-singular matrix P such that P T SP is diagonal with diagonal entries ±1 or. (c What is the type of the SBF on R 2 given by the matrix S? What is its rank and what is its signature? ( ( 2/ 5 1/ 5 [Ans: (a 1/ 5 2/ 2/5 1/5, (b 5, (c (1,1, 2, ] 2/1 2/5 (a Eigenvalues given by (7 λ( 2 λ 36 ( =, whence λ = 1, 5. For λ = 1 an 2 eigenvector is given by 3x 1 + 6x 2, e.g.. For an orthogonal matrix this 1 must be normalised, by dividing by 5. This gives the first column of the matrix given, and the other eigenpair gives the second. (b Divide each column by λ to get the stated matrix. (c At this point the form is represented by the matrix diag(1, 1, which leads to the stated results. 6. Let V denote the vector space of n n real matrices. (a What is the dimension of V and of the subspace of symmetric matrices and of the subspace of antisymmetric matrices? n n (b Let b(x, Y = Trace(XY. Show that b(x, Y = X jk Y kj, and that b defines an SBF on V. j=1 k=1 (c Show that b is positive definite on the subspace of symmetric matrices and negative definite on the subspace of antisymmetric matrices. (d Find the type, rank and signature of b. [Ans: (a n 2, n(n + 1/2, n(n 1/2 (d (n(n + 1/2, n(n 1/2, n 2, n].

5 (a V is spanned by the n 2 matrices V IJ, where (V IJ ij = 1 if i = I and j = J, otherwise. These are obviously linearly independent, and so dim V = n 2. A less formal but more straightforward way of seeing this is that n n matrices have n 2 elements which can be chosen independently. For symmetric matrices the elements above the diagonal determine those below the diagonal; in addition the elements on the diagonal may be chosen independently. Therefore the total number of elements that may be chosen independently is n = n(n + 1/2. For antisymmetric matrices the diagonal entries are, and again those above the diagonal determine those below. Therefore the number of entries which may be chosen independently is less than for symmetrical matrices by the number on the diagonal, n, giving the stated result. (b (XY ii = n j=1 X ijy ji ; hence result. (Recall that the trace is the sum of diagonal entries. b is obviously linear in the first argument. For symmetry we note that n n j=1 k=1 X jky kj = n j=1 Y kjx jk, which is the expression for b(y, X, except that dummy indices n k=1 of summation j, k are interchanged. n k=1 X jkx kj = n j=1 n k=1 X2 jk,. Choose bases of the subspaces of symmetric and antisymmetric matrices with respect to which the matrix of b is diagonal in standard form (Q.4a. Their union is clearly a basis of V with respect to which b is diagonal with n(n + 1/2 values of 1 on the diagonal and n(n 1/2 values of -1 on the diagonal. These are the values of p, q, from which the other results are immediate. (c Let Y = X, where X is symmetric. Then b(x, X = n j=1 by symmetry. This is positive definite (i.e. strictly positive, unless X =. Similarly, if X is antisymmetric we have b(x, X = n j=1 n k=1 X2 jk ( 7. Let b be the SBF on R 4 I2 given by the matrix B given in block form as B =. { ( I 2 Av Let A be a fixed 2 2 matrix. Define U := x R 4 x =, v R }. 2 (We are using v block form notation above. Show that U is a subspace of R 4 and state its dimension. Show that b is identically zero on U if and only if A is an orthogonal matrix (i.e. iff A T A = I. [Ans: 2] ( Let (v 1, v 2 be a basis of R 2. Then clearly U is spanned by the two vectors Avi, i = 1, 2, which are obviously linearly independent. Hence they are a basis of U, v i whose dimension must be 2. ( Av Next, with u =, we have v b(u, u = u T Bu = ( (Av T, v T ( Av B v = ( v T A T, v T( Av v = v T A T Av v T v = v T (A T A Iv

6 Hence b is identically zero iff the form on R 2 defined by the matrix A T A I is identically zero. By considering this form with respect to a basis which puts its matrix in standard form, this happens if and only if A T A I =.

7 8. Consider the quadratic form β = 2x 2 + 3y 2 + z 2 4xy + 2xz + 2yz. (a Write down the (symmetric matrix B of this quadratic form. (b By evaluating determinants only, determine the type of this form. (c Find the eigenvalues and eigenvectors of B and check that the signs of the eigenvalues are consistent with derivation of the type in the previous part. (You might want to use Maple to find the eigenvalues use evalf(linearalgebra[eigenvalues](b. [Ans. (a, 1, 2] , (b (2,1, (c 4.57, -.71, 2.14 or (harder 2 3 sin (a See above. The diagonal entries are the coefficients of x 2, y 2, z 2, and the off-diagonal terms are half the coefficients of the mixed terms (e.g. 4xy give (B 12 = (B 21 = 2. (b Computing 1 1, 2 2 and 3 3 (subdeterminants starting from the top left, the results are 2, 2, 7. Thus the number of changes of sign in the sequence 1, 2, 2, 7 is 1, and so q = 1. The form is non-degenerate (det B and so p = 2. (c Suitable Maple code is >with(linearalgebra: >A:=Matrix([[2,-2,1],[-2,3,1],[1,1,1]]: evalf(eigenvalues(a; ( ( 1 3 arcsin kπ Without evalf the expressions are suprisingly complex, even though it is known that the result is real. Even the numerical evaluation leads to a small complex part, because of rounding error. The trigonometric form given in the answer is obtained by a standard method; see, for example, starting at about eq.( What is the type of the SBF with matrix a 3 3 determinant (or compute eigenvalues - don t even think of it. [Ans: (2,1]? You do not need to evaluate The lower right 1 1 and 2 2 subdeterminants are positive, and therefore b, restricted to the span of the last two basis vectors, is positive definite. Hence p 2. It is clearly negative definite on the span of the first basis vector, and so q 1. But p + q 3 and so the result follows. 1. What is the type of the SBF with matrix (a (b (c

8 [Ans: (a (2,1, (b (2,1, (c (2,2] (a Evaluating subdeterminants, starting from the lower right, we get the sequence 2, 1, 1. There is one sign change (starting with a supplementary 1, hence result. (b Evaluating subdeterminants, starting from top left, gives the sequence 1, 37, 46, with only one sign change. Hence q = 1. (c The top left 1 1 and 2 2 subdeterminants are 3, 2, and so b is negative definite on the span of the first two basis vectors. Thus q 2. Similarly, however, on the span of the last two basis vectors, b is positive definite, and so p 2. But p+q 4, hence result. 11. Classify the following quadrics. (a x 2 + 2y 2 + 3z 2 + 2xy + 2xz = 1 (b 2xy + 2xz + 2yz = 1 (c x 2 + 3y 2 + 2xz + 2yz 6z 2 = 1 You may be able to do all of these using determinants if you think carefully. You can always check your answer by asking Maple for the eigenvalues. [Ans: (a ellipsoid, (b hyperboloid of two sheets (c hyperboloid of one sheet] (a The corresponding matrix is 1 2, and the sequence of subdeterminants 1 3 from top left is 1, 1, 1. No sign change, therefore type (3,. Hence result. 1 1 (b The matrix is 1 1, which looks very tricky. But the form is clearly not 1 1 negative definite, and so p 1 and q 2. Looking at the top left determinant ( 1 we see that, restricted to the x, y plane, the form has type (1,1, and so q 1. Similarly, when restricted to each of the three coordinate planes, it has this type. Hence the subspace on which it is negative definite must intersect these planes, and must itself be a plane (no line can do so. Therefore q = (c The matrix is sign change, and so q = 1., and the sequence of determinants is 1, 3, 22. One 12. Let β(x be a quadratic form on R n given by a symmetric matrix S. How are the maximum and minimum values of β(x on the unit sphere x T x = 1 related to the eigenvalues of S? (Hint: orthogonal change of coordinates to standard form. Continuing, use Lagrange multipliers to find the critical points of x T Sx subject to the constraint x T x = 1. [Ans: max i λ i, min i λ i ; x i (normalised eigenvector]

9 There is an orthogonal change of coordinates to a set of coordinates x such that the matrix of β is diag(λ 1,...,λ n. Also, because the transformation is orthogonal, the unit sphere is x T x = 1. Thus β = λ 1 x λ nx 2 n min(λ i (x x 2 i n = min(λ i, i and the minimum is attainable, by taking x i = 1, where i is the index of the smallest eigenvalue. Similarly for the maximum. The above calculation essentially gives the answer( to the second part of the question. Using Lagrange multipliers, however, we set x i S ij x j µ x 2 i =. Thus x k i,j i S kj x j + x i S ik 2µx k =. By symmetry of S, this is S kj x j + x i S ki 2µx k =, j i j i where the two sums differ only in the dummy index of summation. Hence S kj x j µx k =. j In matrix notation this is just Sx = µx, and so x is an eigenvector. It must be normalised to satisfy the constraint x T x = Consider ( the SBFs on ( R 2 given (with respect to the standard basis by the matrices B =, A =. Show that one of these is positive definite and hence find a basis for R 2 with respect to which the matrices of both the SBFs are diagonal (with the positive definite one having the identity as its matrix. Write down the change of basis matrix that diagonalises both forms. ( 1 [Ans: ] 1 1 The 1 1 and 2 2 leading subdeterminants of B are 1,-6, and so it is not positive definite. For A we get 2,1, and it is positive definite. ( 1 2λ 3 λ det(b λa = det 3 λ 3 λ = (1 2λ(3 λ (3 λ 2, and so λ = 2, 3. For λ = ( 2 (relative eigenvectors are given by (1 2( 2x 1 + (3 a ( 2x 2 =, and so x = is a possibility. It has to be normalised with respect a to A, ( i.e. x T Ax = 1, and so a 2 = 1. We choose a = 1. Similarly for λ = 3 we may get. Hence (in a different order we obtain the columns of the transformation 1 matrix X as given above. It is easy to verify that X T AX and X T BXare diag(1, 1 and diag(3, 2. A couple more Quotient Spaces questions

10 14. Let T be the linear map on R 3 with matrix relative to the standard basis (e 1, e 2, e 3. Show that V = Sp(e 2 is invariant under T (i.e. T(V V. Write down a basis of R 3 /V, and the matrix of the canonical map T with respect to this basis. ( 1 1 [Ans: ([e 1 ], [e 3 ], ] 1 T(e 2 = so V is invariant. Therefore T exists. 1 = 2 = 2e 2. Thus T(e 2 V, and The vectors [e 1 ], [e 3 ] are linearly independent, as a 1 [e 1 ]+a 3 [e 3 ] = [] a 1 e 1 +a 3 e 3 V, i.e. there exists a 2 such that a 1 e 1 + a 3 e 3 = a 2 e 2. Since (e 1, e 2, e 3 is a basis of R 3, a 1 = a 2 = a 3 =. Also dim R 3 /V = dim R 3 dim V = 2, and so ([e 1 ], [e 3 ] must be a basis of R 3 /V. T([e 1 ] = [T(e 1 ] = [e 1 + e 3 ] = [e 1 ] + [e 3 ] and T([e 3 ] = [T(e 3 ] = [ e 1 ] = [e 1 ]. The coefficients of [e 1 ], [e 3 ] give the columns of the required matrix. 15. Consider the differentiation map D : P 3 P 3, where P 3 is the real vector space of polynomials of degree 3 in a variable x. Show that D gives rise to a linear map D : P 3 /V P 3 /V, where V is the subspace of constant polynomials. What is the matrix of D with respect to the basis ([x], [x 2 ], [x 3 ] of P 3 /V? [Ans: 2 3 ] Since V is invariant under D, D is well defined. Then D([x] = [D(x] = [1] = P 3 /V D([x 2 ] = [D(x 2 ] = [2x] = 2[x] D([x 3 ] = [D(x 3 ] = [3x 2 ] = 3[x 2 ] from which the columns of the matrix are immediate.