Exercises on similar matrices and Jordan form Problem 28.1: (6.6 #12. Introduction to Linear Algebra: Strang) These Jordan matrices have eigenvalues 0, 0, 0, 0. They have two eigenvectors; one from each block. However, their block sizes don’t match and they are not similar: ⎡ ⎤ ⎡ ⎤ 0 1 0 0 0 1 0 0 ⎢ 0 0 0 0 ⎥ ⎢ ⎥ ⎥ and K = ⎢ 0 0 1 0 ⎥ . J=⎢ ⎣ 0 0 0 1 ⎦ ⎣ 0 0 0 0 ⎦ 0 0 0 0 0 0 0 0 For a generic matrix M, show that if J M = MK then M is not invertible and so J is not similar to K. Solution: Let M = (mij ). Then: ⎡ ⎤ ⎡ m21 m22 m23 m24 0 m11 ⎢ ⎥ ⎢ 0 0 0 0 ⎥ ⎢ 0 m21 JM = ⎢ ⎣ m41 m42 m43 m44 ⎦ and MK = ⎣ 0 m31 0 0 0 0 0 m41 m12 m22 m32 m42 ⎤ 0 0 ⎥ ⎥ 0 ⎦ 0 If J M = MK, then m11 = m22 = 0, m21 = 0, m31 = m42 = 0, and m41 = 0. Thus, the ﬁrst column of M is all zeros and M is not invertible. If J were similar to K there would be an invertible matrix M that satis ﬁes K = M−1 J M, and so MK = J M. We just showed that there can be no such invertible matrix M. Therefore J is not similar to K. Problem 28.2: (6.6 #20.) Why are these statements all true? a) If A is similar to B then A2 is similar to B2 . b) A2 and B2 can be similar when A and B are not similar (try λ = 0, 0.) � � � � 3 0 3 1 c) is similar to . 0 4 0 4 � � � � 3 0 3 1 d) is not similar to . 0 3 0 3 e) Given a matrix A, let B be the matrix obtained by exchanging rows 1 and 2 of A and then exchanging columns 1 and 2 of A. Show that A is similar to B. Solution: a) If A is similar to B, then: A = M−1 BM =⇒ A2 = M−1 BM( M−1 BM) = M−1 B2 M. Since A2 = M−1 B2 M, by deﬁnition A2 is similar to B2 . b) Let: � A= 0 0 0 0 � � and B = 0 1 0 0 � . Then A2 = B2 and so A2 is similar to B2 , but A is not similar to B because nothing but the zero matrix is similar to the zero matrix. c) There are multiple ways to verify that the given matrices are similar. One way is to explicitly ﬁnd the matrix M that satisﬁes the similarity 3 0 condition: � 1 −�1 = 4 0 0 1 � 3 1 0 4 � �1 1 0 1 �� . � d) The given matrices are not similar because for the ﬁrst matrix every vector in the plane is an eigenvector for λ = 3, whereas the second matrix only has a line (y = 0) of eigenvectors corresponding to λ = 3. We know that similar matrices have the same number of independent eigenvectors, so the given matrices cannot be similar. e) To exchange the ﬁrst two rows of A, we multiply A on the left by: ⎡ ⎤ 0 1 0 0 ⎢ 1 0 0 0 ⎥ ⎥ M=⎢ ⎣ 0 0 1 0 ⎦, 0 0 0 1 In order to exchange the ﬁrst two columns of A, we multiply it on the right by the same matrix M. We thus have B = MAM. Since M−1 = M, B is similar to A.