Linear Algebra

352 Chapter Five. SimilarityIn the x = 1 possibility, the first equation in (∗) is 2 · b 1 + 2 · b 2 = 0, and soassociated with 1 are vectors whose second component is the negative of theirfirst component. ( ) ( ) ( )3 2 b1 b1= 1 ·0 1 −b 1 −b 1Thus, another solution is λ 2 = 1 and a second basis vector is this.(1 ⃗β 2 =−1)To finish, drawing the similarity diagramR 2 w.r.t. E 2⏐id↓R 2 w.r.t. Bt−−−−→TR 2 w.r.t. E 2t−−−−→D⏐id↓R 2 w.r.t. Band noting that the matrix Rep B,E2 (id) is easy leads to this diagonalization.( ) ( ) −1 ( ) ( )3 0 1 1 3 2 1 1=0 1 0 −1 0 1 0 −1In the next subsection, we will expand on that example by considering moreclosely the property of Corollary 2.4. This includes seeing another way, the waythat we will routinely use, to find the λ’s.Exerciseš 2.6 Repeat Example 2.5 for the matrix from Example 2.2.2.7 Diagonalize these upper triangular matrices.(a)(−2 10 2)(b)(5 40 1)̌ 2.8 What form do the powers of a diagonal matrix have?2.9 Give two same-sized diagonal matrices that are not similar. Must any twodifferent diagonal matrices come from different similarity classes?2.10 Give a nonsingular diagonal matrix. Can a diagonal matrix ever be singular?̌ 2.11 Show that the inverse of a diagonal matrix is the diagonal of the the inverses,if no element on that diagonal is zero. What happens when a diagonal entry iszero?2.12 The equation ending Example 2.5(1 10 −1) −1 ( ) ( )3 2 1 1=0 1 0 −1( )3 00 1is a bit jarring because for P we must take the first matrix, which is shown as aninverse, and for P −1 we take the inverse of the first matrix, so that the two −1powers cancel and this matrix is shown without a superscript −1.(a) Check that this nicer-appearing equation holds.(3 00 1)=(1 10 −1) ( ) (3 2 1 10 1 0 −1) −1