ALGEBRA US02CMTH21 MATRIX THEORY Shreya Chauhan April 21, 2021 1 1 Special Types of Matrices Zero Divisors: If A and B are nonzero matrices such that AB = 0 then A and B are called zero divisors. 2 3 2 3 2 3 1 0 0 0 0 0 e.g. A = 4 5 B = 4 5 Then AB = 4 5 0 0 1 0 0 0 Theorem: The product of two matrices can be a zero matrix though none of them is a zero matrix. 2 3 2 3 0 c −b a2 ab ac 6 7 6 7 6 7 6 2 7 Let A = 6 −c 0 a 7 And B = 6 ab b bc 7 4 5 4 5 b −a 0 ac bc c2 Then 2 3 abc − abc b2c − b2c bc2 − bc2 6 7 6 2 2 2 2 7 AB = 6 −a c + a c −abc + abc −ac + ac 7 4 5 ba2 − a2b ab2 − ab2 abc − abc 2 3 0 0 0 6 7 6 7 = 6 0 0 0 7 4 5 0 0 0 while A 6= 0 B 6= 0 Important Result: 1 AB = AC does not imply B = C. i.e. cancellation laws does not hold for matrix multiplication. 2 3 2 3 2 3 1 0 0 0 0 0 e.g. Let A = 4 5 B = 4 5 C = 4 5 0 0 1 0 0 1 Then 2 3 0 0 AB = 4 5 0 0 And 2 3 0 0 AC = 4 5 0 0 But B 6= C. Idempotent Matrix: A square matrix A is said to be idempotent if A2 = A. For example, 2 3 1 0 0 6 7 6 7 A = 6 0 1 0 7. 4 5 0 0 1 Then A2 = A. Hence A is an idempotent matrix. A square matrix A is said to be idempotent of period p if p is the least positive integer such that Ap+1 = A. 2 For example, 2 3 −1 0 A = 4 5 0 −1 Then 2 3 1 0 A2 = 4 5 0 1 And 2 3 2 3 1 0 −1 0 A3 = 4 5 4 5 0 1 0 −1 2+1 So A = A. ) A is an idempotent of period 2. Nilpotent Matrix: A square matrix A is said to be a nilpotent matrix if Ak = 0 where k is a positive integer. If however k is the least integer for which Ak = 0 then k is called the index of nilpotent matrix A. For example, 2 3 0 0 1 6 7 6 7 A = 6 0 0 1 7 4 5 0 0 0 Then 3 2 3 0 0 0 6 7 2 6 7 A = 6 0 0 0 7 4 5 0 0 0 ) A is a nilpotent matrix index 2. Involuntary Matrix: A square matrix A is said to be an innvolutary matrix if A2 = I. Since I2 = I always, hence unit matrix I is involuntary matrix. 2 3 −1 0 Let A = 4 5. 0 −1 2 3 1 0 Then A2 = 4 5. 0 1 Hence A is also involuntary matrix. Orthogonal Matrix: A square matrix A is said to be orthogonal if AA0 = I = A0A. For example, 2 3 0 1 A = 4 5 1 0 Then 2 3 0 1 A0 = 4 5 1 0 4 Then 2 3 2 3 0 1 0 1 AA0 = 4 5 4 5 1 0 1 0 2 3 1 0 = 4 5 0 1 Similarly we can show that A0A = I. Hence A is an orthogonal matrix. Unitary Matrix: A square matrix A is said to be unitary if AθA = I = AAθ. For example, 2 3 0 i A = 4 5 i 0 Then 2 3 0 −i Aθ = 4 5 −i 0 Then 2 3 2 3 0 i 0 −i AAθ = 4 5 4 5 i 0 −i 0 2 3 −i2 0 = 4 5 0 −i2 2 3 1 0 = 4 5 0 1 5 Similarly we can show that AθA = I. Hence A is a unitary matrix. Theorem 1: If A and B are two idempotent matrices and if AB = BA = 0 then A + B will also be an idempotent matrix. Proof: Let A and B be two idempotent matrices. Then A2 = A and B2 = B, and suppose AB = BA = 0. Now (A + B)2 = (A + B)(A + B) = A(A + B) + B(A + B)(* DistributiveLaw) 2 2 = A + AB + BA + B (* DistributiveLaw) 2 2 = A + B (* A = A and B = B) Hence A + B is an idempotent matrix. Theorem 2: If A and B are two idempotent matrices and if they com- mute then AB is idempotent matrix. Proof: Let A and B be two idempotent matrices. Then A2 = A and B2 = B. Assume the A and B commutes. i.e. AB = BA 6 Now (AB)2 = (AB)(AB) = A(BA)B (* AssociativeLaw) = A(AB)B (* AB = BA) = (AA)(BB)(* AssociativeLaw) = A2B2 = AB Hence AB is idempotent matrix. Theorem 3: If A and B are two n-rowed orthogonal matrices then AB and BA are also orthogonal matrices. Proof: Let A and B be two orthogonal matrices. Then AA0 = A0A = I And BB0 = B0B = I Now (AB)(AB)0 = (AB)(B0A0) 0 0 = A(BB )A (* AssociativeLaw) = A(I)A0 = (AI)A0 = AA0 = I 7 Similarly (AB)0(AB) = (B0A0)(AB) 0 0 = B (A A)B (* AssociativeLaw) = B0(I)B = (B0I)B = B0B = I So we get (AB)(AB)0 = I = (AB)0(AB). Hence AB is an orthogonal matrix. Similarly we can show that BA is an orthogonal matrix. Theorem 4: If A and B are two n-rowed unitary matrices then AB and BA are also unitary matrices. Proof: Let A and B be two unitary matrices. Then AAθ = AθA = I And BBθ = BθB = I Now (AB)(AB)θ = (AB)(BθAθ) θ θ = A(BB )A (* AssociativeLaw) = A(I)Aθ = (AI)Aθ = AAθ = I 8 Similarly (AB)θ(AB) = (BθAθ)(AB) θ θ = B (A A)B (* AssociativeLaw) = Bθ(I)B = (BθI)B = BθB = I So we get (AB)(AB)θ = I = (AB)θ(AB). Hence AB is an unitary matrix. Similarly we can show that BA is an unitary matrix. 2 3 2 −2 −4 6 7 6 7 Example 1: Show that the matrix A = 6 −1 3 4 7 is an idem- 4 5 1 −2 −3 potent matrix. Solution: We have 2 3 2 −2 −4 6 7 6 7 A = 6 −1 3 4 7 4 5 1 −2 −3 9 Then 2 3 2 3 2 −2 −4 2 −2 −4 6 7 6 7 2 6 7 6 7 A = 6 −1 3 4 7 6 −1 3 4 7 4 5 4 5 1 −2 −3 1 −2 −3 2 3 4 + 2 − 4 −4 − 6 + 8 −8 − 8 + 12 6 7 6 7 = 6 −2 − 3 + 4 2 + 9 − 8 4 + 12 − 12 7 4 5 2 + 2 − 3 −2 − 6 + 6 −4 − 8 + 9 2 3 2 −2 −4 6 7 6 7 = 6 −1 3 4 7 4 5 1 −2 −3 = A 2 ) A = A. Hence A is idempotent matrix. 2 3 1 1 3 6 7 6 7 Example 2: Show that 6 5 2 6 7 is a nilpotent matrix of index 3. 4 5 −2 −1 −3 10 2 3 1 1 3 6 7 6 7 Solution: Let A = 6 5 2 6 7 4 5 −2 −1 −3 Then 2 3 2 3 1 1 3 1 1 3 6 7 6 7 2 6 7 6 7 A = 6 5 2 6 7 6 5 2 6 7 4 5 4 5 −2 −1 −3 −2 −1 −3 2 3 1 + 5 − 6 1 + 2 − 3 3 + 6 − 9 6 7 6 7 = 6 5 + 10 − 12 5 + 4 − 6 15 + 12 − 18 7 4 5 −2 − 5 + 6 −2 − 3 + 3 −6 − 6 + 9 2 3 0 0 0 6 7 6 7 = 6 3 3 −3 7 4 5 −1 −1 −3 11 Then A3 = A2A 2 3 2 3 0 0 0 1 1 3 6 7 6 7 6 7 6 7 = 6 3 3 −3 7 6 5 2 6 7 4 5 4 5 −1 −1 −3 −2 −1 −3 2 3 0 0 0 6 7 6 7 = 6 0 0 0 7 4 5 0 0 0 Thus A3 = 0. Hence A is a nilpotent matrix of index 3. 2 3 1 1 + i Example 3: Prove that the matrix p1 4 5 is unitary. 3 1 − i −1 2 3 1 1 + i Solution: Let A = p1 6 7 3 4 5 1 − i −1 2 3 1 1 − i Then A0 = p1 4 5 3 1 + i −1 2 3 1 1 + i Then Aθ = A0 = p1 4 5 3 1 − i −1 12 Then 2 3 2 3 θ 1 1 1 + i 1 1 1 + i AA = p 4 5 p 4 5 3 1 − i −1 3 1 − i −1 2 3 1 1 + (1 − i)2 1 + i − (1 + i) = 4 5 3 (1 − i) − (1 + i) 1 − i2 + 1 2 3 1 3 0 = 4 5 3 0 3 2 3 1 0 = 4 5 0 1 = I Thus AAθ = I. Similarly we get AθA = I Hence A is a unitary matrix. 2 3 −5 −8 0 6 7 6 7 Example 4: Show that A = 6 3 5 0 7 is involuntary.