Linear Algebra

Section II. Similarity 363b 2 = 0 case, the first equation gives that either b 1 = 0 or x = 3. Since we’vedisallowed the case of both b 1 = 0 and b 2 = 0, we are left with λ 1 = 3. Thenthe first equation in (∗) is 0 · b 1 + 2 · b 2 = 0 and so associated with λ 1 = 3 arevectors with a second component of zero and a first component that is free.( ) ( ) ( )3 2 b 1 b 1= 3 ·0 1 0 0Choose any nonzero b 1 to have a first basis vector.( )1⃗β 1 =0The second case for the bottom equation of (∗) is λ 2 = 1. The first equation in(∗) is then 2 · b 1 + 2 · b 2 = 0 and so associated with 1 are vectors such that theirsecond component is the negative of their first.( ) ( ) ( )3 2 b 1 b 1= 1 ·0 1 −b 1 −b 1Choose a nonzero one of these to have a second basis vector.( )1⃗β 2 =−1Now drawing the similarity diagramR 2 wrt E 2id⏐↓R 2 wrt Bt−−−−→TR 2 wrt E 2t−−−−→Did⏐↓R 2 wrt Band noting that the matrix Rep B,E2 (id) is easy gives us 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 Lemma 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.