Linear algebra and matrices M3P9

Question Sheet 7 - Answers and solutions

1. (1) The transpose of A.A^{−1} = Id is (A^{−1})^{t}.A^{t} = Id, hence the transpose of A^{−1} is the inverse to A^{t}, that is, (A^{−1})^{t} = (A^{t})^{−1}. But this is just A^{−1} since A is symmetric.

(2) We have u^{t}.A.A^{t}.v = u^{t}.(A^{t})^{t}.A^{t}.v = (A^{t}.u)^{t}.(A^{t}.v). It is now clear that the bilinear form < u, v >:= u^{t}.A.A^{t}.v is symmetric. If u = v we have < u, u >= w^{t}.w where w = A^{t}u. If u ≠ 0, then w ≠ 0 (since A^{t} is invertible), and then w^{t}.w = ∑_{i=1}^{n} w_{i}^{2} is positive.

(3) There exists an orthogonal matrix P such that P^{t}.A.P is diagonal, with some real λ_{1}, …, λ_{n} on the diagonal. At the same time P^{t}.Id.P = P^{t}.P = Id (since P is orthogonal). Hence for any real c the matrix P^{t}(A + c.Id)P is diagonal with λ_{1} + c, …, λ_{n} + c on the diagonal. If c is big enough, it has only positive entries in the main diagonal, hence defines a positively definite form.

2. (1) By sheet 6, the eigenvalues of A are complex numbers of the form y\sqrt{−1}, hence the eigenvalues of Id + A are of the form 1 + y\sqrt{−1}. Thus 0 is not an eigenvalue of Id + A, hence this matrix is invertible.

(2) We have ((Id − A)(Id + A)^{−1})^{t} = (Id + A^{t})^{−1}(Id − A^{t}) = (Id − A)^{−1}(Id + A) (here we used the fact the operations of taking the transpose and taking the inverse commute, see Question 1). Next, a matrix commutes with its inverse, thus the above expression equals (Id + A)(Id − A)^{−1} which is clearly the inverse of (Id − A)(Id + A)^{−1}.

3. Let Az = λ.z, where z = (z_{1}, …, z_{n})^{t} ≠ 0, z_{i} ∈ C, λ ∈ C. Then \overline{z}^{t}.z = \overline{z}^{t}.A^{t}.A.z = \overline{Az}^{t}.Az (since A is real and orthogonal), but this also equals \overline{λ}λ.\overline{z}^{t}.z. Since \overline{z}^{t}.z > 0 we must have |λ| = 1.

4. Suppose that a hyperboloid of 2 sheets H, say given by ax^{2} + by^{2} = −1 + cz^{2}, a, b, c > 0, contains at least 3 distinct points of a straight line L given by x = x_{0} + x_{1}t, y = y_{0} + y_{1}t, z = z_{0} + z_{1}t. Then the quadratic polynomial

(ax_{1}^{2} + by_{1}^{2} − cz_{1}^{2})t^{2} + 2(ax_{0}x_{1} + by_{0}y_{1} − cz_{0}z_{1})t + (ax_{0}^{2} + by_{0}^{2} − cz_{0}^{2} + 1) = 0 has at least 3 distinct roots, therefore all its coefficients are zeros. Thus L ⊂ H. (Note that this argument applies to any quadric surface!) Next, the section of H by the plane z = 0 is empty, hence L must be parallel to this plane, that is, z_{1} = 0, z_{0} ≠ 0. But the section of H by the plane z = z_{0} is an ellipse, a point, or an empty set, hence contains no lines.

5. The answer is “no”. A hyperbolic paraboloid P can be given by ax^{2} − by^{2} = z, a, b > 0. Suppose that we have a line L ⊂ P in the same parametrical form as in the previous exercise. Then ax_{1}^{2} − by_{1}^{2} = 0. It follows from this equation that if x_{1} = 0, then y_{1} = 0. But then the LHS of ax^{2} − by^{2} = z after restriction to L (substituting x, y, z as functions in t) is a constant, hence so is z = z_{0} + z_{1}t, but then (x_{1}, y_{1}, z_{1}) = (0, 0, 0) which is not possible. We can thus assume without loss of generality that x_{1} = 1, and then we can assume that x = t (by rerplacing the parameter t by t + x_{0}). The line L takes the form: y = y_{0} + y_{1}x, z = z_{0} + z_{1}x, and y_{1} = ±\sqrt{a/b}. Substituting these into the equation of P we obtain z_{1} = −2(±\sqrt{ab})y_{0}, z_{0} = −by_{0}^{2}. This shows that the triple (x_{1}, y_{1}, z_{1}) = (1, ±\sqrt{a/b}, z_{1}) uniquely determines (0, y_{0}, z_{0}) (for each choice of sign), and hence determines the line.

If L′ is parallel to L, then it has the same direction vector (1, y_{1}, z_{1}). However, we have just seen that it uniquely determines the line in P with this direction vector. Thus L′ = L.