Back to Invertibility

Exercise 10

Avg difficulty 2.5

We say that two square matrices AA and BB are similar if there is an invertible matrix PP such that B=PAP−1B = PAP^{-1}.

  1. If AA and BB are similar, explain why A2A^2 and B2B^2 are similar as well. In particular, if B=PAP−1B = PAP^{-1}, explain why B2=PA2P−1B^2 = PA^2P^{-1}.

  2. If AA and BB are similar and AA is invertible, explain why BB is also invertible.

  3. If AA and BB are similar and both are invertible, explain why A−1A^{-1} and B−1B^{-1} are similar.

  4. If AA is similar to BB and BB is similar to CC, explain why AA is similar to CC. To begin, you may wish to assume that B=PAP−1B = PAP^{-1} and C=QBQ−1C = QBQ^{-1}.

Austin, Understanding Linear Algebra, §3.1 Invertibility, Exercise 10

AI Tutor

Adapted from Understanding Linear Algebra © 2023 David Austin. Licensed under CC BY 4.0. Source: Understanding Linear Algebra. Imported at revision 3e6dced.