Alg1231W8T3 - Diagonalisation

Wednesday, 12 September 2018

10:12 PM

Machine generated alternative text: If we have a matrix A and we construct a matrix M , whose columns are just eigenvectors of A , then we always have AM = MD for some diagonal matrix D However, if we wish to use this matrix to write down a nice factored form for A, we must take care when constructing M to ensure that M is an invertible matrix. -11 Consider the matrix A 7 If we define -14 This has eigenvectors VI — 10 and v 2 AVI = 3 VI and where AV2 — —4". 10 —10 (i.e define L to be the matrix whose columns are VI and 10V1)v then we can calculate 1104) ( —11 Note: the Maple notation for the matrix 112 >,< 314 3 o 10 -10 3 Since the columns of L are linearly dependent; the matrix L If we instead define 10 -10 whose columns are the linearly independent eigenvectors above, then we get —11 AM is not invertible! So we cannot use this to diagonalise A. 1104) ( 3

 

Machine generated alternative text: Since the columns of M are linearly independent, the matrix M is invertible. Namely M has inverse By multiplying on the right by M we get A = MDM¯I where In this case A is said to be diagonalisable

 

Machine generated alternative text: Two eigenvectors with different eigenvalues are always linearly independent To see why, take a matrix A with eigenvectors VI and v2 such that AVI = AV2 = tv,2 where s # t. Recall that VI and v2 must be non-zero vectors, since being non-zero is part of the definition of an eigenvector Now consider the equation PVI + qu = O. (1) lution to (1) isp= q = O , then we will have shown VI and v2 are linearly independent If we can show that the only Since PVI = —qv,2 we have spvl = A(pvl) = A(—qv2) —tqv2 = tPV1 • Then, by rearrangement p(s — t)V1 = O. There are three cases to consider: VI = O, which is impossible since eigenvectors are non-zero by definition; O @twhich is impossible since the eigenvalues were assumed to be different, • orp=0. Only one of these three cases is possible, so we conclude p = 0_ Then equation (1) becomes qv2 = O. Since v2 must be a non-zero vector the only way to solve this is to have q = 0. Thus we have show that v2 must be linearly independent. the only solution O IS p 0, and hence the vectors VI and

 

Machine generated alternative text: The number of linearly independent eigenvectors for the eigenvalue is equal to nullity(A — M). • if A then there is one eigenvalue = 2 and nullity(A — 21) = 2 • ifB then there is only one eigenvalue = 2 and nullity(B — 21) = 1 Hence; O and the number of linearly independent eigenvectors for = 2 is 2 O and the number of linearly independent eigenvectors for = 2 is 1 Because the matrix B does not have enough linearly independent eigenvectors, it is called a defective matrix, and cannot be diagonalised

 

Machine generated alternative text: Which of the following matrices are diagonalisable? 4 10 any n x n matrix with n linearly independent eigenvectors any 2 x 2 matrix with two distinct eigenvalues

 

Machine generated alternative text: 1 1 We are told that A has eigenvalues 5 and 3 with respective eigenvectors -2 and . So we can diagonalise A by finding an invertible matrix —1 and diagonal matrix D such that 5 i) If we wanted D to be the matrix D o M-IAM then we should take 3 ii) If instead we wanted D to be the diagonal matrix D then we should take iii) One of the advantages of having a diagonal matrix is that matrix powers are easy to take, for example 6 The relation for ATI is not nearly as simple. iv) Once we write A = MDM¯I we observe that 142 MD2M¯1 .43 = MD3M¯1, and so on. Thus, using parts (i) and (iii), we can easily calculate that the top left entry of 146 is 30521 o. 1 Recall: the Maple notation for the matrix 3 2 is 4

 

 

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