Exercises on symmetric matrices and positive deﬁniteness Problem 25.1: (6.4 #10. Introduction to Linear Algebra: Strang) Here is a quick “proof” that the eigenvalues of all real matrices are real: False Proof: Ax = λx gives x T Ax = λx T x so λ = xT Ax is real. xT x There is a hidden assumption in this proof which is not justiﬁed. Find the ﬂaw by testing each step on the 90 ◦ rotation matrix: � � 0 −1 1 0 with λ = i and x = (i, 1). � Solution: We can esily conﬁrm that Ax = λx = −1 i � . Next, check if xT Ax = λxT x is true for the 90 ◦ rotation matrix: � � � � −1 T i 1 x Ax = =0 i � � � � i T λx x = i i 1 =0 1 x T Ax = λx T x.� Note that xT x = 0. Since the next and last step involves dividing by this term, the hidden assumption must be that xT x �= 0. If x = ( a, b) then � � � � a T x x= a b = a2 + b2 . b The “proof” assumes that the squares of the components of the eigenvec tor cannot sum to zero: a2 + b2 �= 0. This may be false if the components are complex. Problem 25.2: (6.5 #32.) A group of nonsingular matrices includes AB and A−1 if it includes A and B. “Products and inverses stay in the group.” Which of these are groups? a) Positive deﬁnite symmetric matrices A. b) Orthogonal matrices Q. c) All exponentials etA of a ﬁxed matrix A. d) Matrices D with determinant 1. Solution: a) The positive deﬁnite symmetric matrices A do not form a group. To show this, we provide a counterexample in the form of two positive deﬁnite symmetric matrices A and B whose product is not a positive deﬁnite symmetric matrix. � � � � � � 2 1 1 1/2 2.5 2 If A = and B = then AB = is not 1 1 1/2 1 1.5 1.5 symmetric. b) The orthogonal matrices Q form a group. If A and B are orthogonal matrices, then: A T A = I ⇒ A−1 = A T ⇒ A−1 is orthogonal, and B T B = I ⇒ ( AB) T AB = B T A T AB = B T B = I ⇒ AB is orthogonal. c) The exponentials etA of a ﬁxed matrix A form a group. For the elements e pA and eqA : (e pA )−1 = e− pA is of the form etA e pA eqA = e( p+q) A is of the form etA d) The matrices D with determinant 1 form a group. If det A = 1 then det A−1 = 1. If matrices A and B have determinant 1 then their product also has determinant 1: det( AB) = det( A) det( B) = 1.