Exercises on diagonalization and powers of A Problem 22.1: (6.2 #6. Introduction to Linear Algebra: Strang) Describe all matrices S that diagonalize this matrix A (ﬁnd all eigenvectors): � � 4 0 A= . 1 2 Then describe all matrices that diagonalize A−1 . Solution: To ﬁnd the eigenvectors of A, we ﬁrst ﬁnd the eigenvalues: � � 4−λ 0 det = 0 =⇒ (4 − λ)(2 − λ) = 0. 1 2−λ Hence the eigenvalues are λ1 = 4 and λ2 = 2. Using these values, we ﬁnd the eigenvectors by solving ( A − λI )x = 0 : � �� 0 0 y 0 ( A − λ1 I ) x = = =⇒ y = 2z, 1 −2 z 0 thus any multiple of (2,1) is an eigenvector for λ1 . � �� 2 0 y 0 ( A − λ2 I ) x = = =⇒ y = 0, z = free variable, 1 0 z 0 thus any multiple of (0,1) is an eigenvector for λ2 . Therefore the columns of the matrices S that diagonalize A are nonzero multiples of (2,1) and (1,0). They can appear in either order. Finally, because A−1 = SΛ−1 S−1 the same matrices S will diagonalize A −1 . Problem 22.2: (6.2 #16.) Find Λ and S to diagonalize A : � � .6 .9 A= . .4 .1 What is the limit of Λk as k → ∞? What is the limit matrix of SΛk S−1 ? In the columns of this matrix you see the . Solution: Since each of the columns of A sums to one, A is a Markov matrix and deﬁnitely has eigenvalue λ1 = 1. The trace of A is .7, so the other eigenvalue is λ2 = .7 − 1 = −.3. To ﬁnd S we need to ﬁnd the corresponding eigenvectors: � �� −.4 .9 y 0 ( A − λ1 I ) x 1 = =⇒ x1 = (9, 4). = .4 −.9 z 0 � �� .9 .9 y 0 = ( A − λ2 I ) x 2 = =⇒ y = −z =⇒ x2 = (1, −1). z 0 .4 .4 Putting these together, we have: � � � � � � 9 1 1 1 k S= and Λ = . As k → ∞, Λ → . 4 −1 −.3 0 So ∞ −1 SΛ S � → 9 1 4 −1 �� 1 0 1 13 �� 1 1 4 −9 � 1 = 13 � In the columns of this matrix you see the steady state vector. 9 9 4 4 � .