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PRACTICING PROOFS This file contains two sets of problems to practice your ability with proofs. Solutions to the first set of problems are provided. The solutions to the second set of problems are intentionally left to the reader (as an incentive to practice!). For your convenience, we begin by recalling some preliminary definitions and theorems that can be used to solve the problems below. PRELIMINARY DEFINITIONS / THEOREMS Definition 1. An n × n matrix B is called idempotent if B 2 = B. Example The identity matrix is idempotent, because I 2 = I · I = I. Definition 2. An n × n matrix B is called nilpotent if there exists a power of the matrix B which is equal to the zero matrix. This means that there is an index k such that B k = O. 0 1 2 Example The zero matrix is obviously nilpotent. The matrix B = 0 0 1 0 0 0 is also nilpotent, because B 3 = B · B · B = O. (Note that any higher power of B is also zero.) Definition 3. An n×n matrix B is called symmetric if it is equal to its transpose: B = B T . An n × n matrix B is called skew-symmetric if B = −B T . 0 1 2 Example The matrix B = 1 0 1 is symmetric. 2 1 3 0 1 2 The matrix B = −1 0 1 is skew-symmetric. −2 −1 0 Definition 4. An n × n matrix B is called non-singular (or “invertible”) if it has a multiplicative inverse, and is called singular (or “not invertible”) otherwise. Theorem 1. For every pair of n × n matrices A and B, det(AB) = det(A)det(B). 1 2 Theorem 2. The following conditions are equivalent: • A is non-singular. • det(A) 6= 0. • The system Ax = 0 has a unique solution (x = 0). Theorem 3. For every n× n matrix A, the determinant of A equals the product of its eigenvalues. PRACTICE PROBLEMS (solutions provided below) (1) Let A be an n × n matrix. Prove that if A is idempotent, then det(A) is equal to either 0 or 1. (2) Let A be an n × n matrix. Prove that if A is idempotent, then the matrix I − A is also idempotent. (Here I is the identity matrix.) (3) Let A be an n × n matrix. Prove that if A is nilpotent, then det(A) = 0. 1 1 (4) Let B be the matrix 0 2 0 0 that the matrix A is invertible 1 1 , and let A be any 3x3 matrix. Prove 3 if and only if the matrix AB is invertible. (5) Let v be any vector of length 3. Let A = (v, 2v, 3v) be the 3 × 3 matrix with columns v, 2v, 3v. Prove that A is singular. (6) Let A = (a1 , a2 , a3 , a4 ) be a 4 × 4 matrix with columns a1 , a2 , a3 , a4 . Suppose that a1 − 3a4 = 0 Prove that A is singular. (the zero vector). 3 ADDITIONAL PROBLEMS (without solutions) (7) Let A be an n × n invertible matrix. Suppose that A−1 = A. Prove that det(A) is equal to either +1 or −1. (8) Suppose that A is an n × n matrix and that 0 is an eigenvalue of A. Prove that A is not invertible. (9) Suppose that A is an n × n matrix and that A2 + 3A = I. Prove that A is invertible. (10) Let A, B, C be n × n invertible matrices. Prove that the product ABC is also invertible. (11) Let A and B be n × n matrices. Suppose that B is invertible and that A = B −1 AB. Prove that A and B commute. (12) Let A be any n × n matrix. Prove that the matrix A + AT is symmetric. (13) Let A be any n×n matrix. Prove that the matrix A−AT is skew-symmetric. (14) Prove that every n × n matrix can be written as the sum of a symmetric matrix and a skew symmetric matrix. 4 Solutions to the first set of practice problems. Problem 1 Suppose that A is idempotent, that is, A2 = A. Taking the determinant of both sides of this equation, we find: det(A2 ) = det(A). (1) Recall that the determinant of two matrices equals the product of the two determinants (see Theorem 1). Then (2) det(A2 ) = det(A · A) = det(A) · det(A) = (det(A))2 . Combining equations (1) and (2), we find that (det(A))2 = det(A). Hence det(A) · [det(A) − 1] = 0, and det(A) = 0 or 1. Problem 2 Suppose that A is idempotent, that is, A2 = A. To prove that the matrix B = I −A is also idempotent, we must show that B 2 = B. Hence, we compute B 2 , and we verify that B 2 is equal to B. B 2 = (I − A)2 = (I − A)(I − A) = A2 = AI + |{z} IA − |{z} = |{z} I 2 − |{z} =I =A =A =A =I −A−A+A= = I − A = B. Note that the only things we used are the definition of idempotent matrix and the fact that multiplication by identity matrix leaves every matrix unchanged. Problem 3 Suppose that A is nilpotent, that is, there exists an index m ≥ 1 such that Am = O. 5 (Here O is the zero matrix.) Taking the determinant of both sides of the equation, we find det(Am ) = det(O). Now, det(O) = 0 and det(Am ) = (det(A))m . (The latter can be proven iterating the argument in problem 1, or, more elegantly by induction.) Then (det(A))m = 0, so det(A) = 0. Problem 4 The matrix B is upper triangular, hence the determinant of B equals the product of the elements on the diagonal: 1 1 1 det(B) = det 0 2 1 = 1 · 2 · 3 = 6. 0 0 3 This gives det(AB) = det(A)det(B) = 6det(A). Then, it’s clear that det(AB) = 0 ⇔ det(A) = 0 and, of course, det(AB) 6= 0 ⇔ det(A) 6= 0. This is equivalent to saying that AB is invertible if and only if A is invertible. Problem 5 It is sufficient to prove that the matrix A = (v, 2v, 3v) has determinant zero. Consider the matrix A′ obtained from A by subtracting twice the first column from the second column: A′ = (v, 2v − 2v, 3v) = (v, 0, 3v). This kind of operation does not affect the determinant, so (3) det(A′ ) = det(A). Because A′ has a zero column, det(A′ ) = 0. Equation (3) implies that det(A) = 0, as well. 6 Problem 6 The matrix A is 4 × 4, with columns a1 , a2 , a3 , a4 . Write A = (a1 , a2 , a3 , a4 ). b1 b2 For every vector b = b3 , we can write: b4 b1 b2 Ab = (a1 , a2 , a3 , a4 ) b3 = b1 a1 + b2 a2 + b3 a3 + b4 a4 . b4 1 0 , we find: In particular, if b = 0 −3 1 1 0 0 A 0 = (a1 , a2 , a3 , a4 ) 0 −3 −3 = 1a1 + 0a2 + 0a3 + (−3)a4 = a1 − 3a4 . By assumption, a1 − 3a4 = 0. Then 1 0 A 0 = 0. −3 1 0 Because the vector 0 is non-zero, this shows that the system Ax = 0 has a −3 non-trivial solution. Hence, A is singular. [See Theorem 2.]