2

$A$ is a non-singular matrix ($n \times n$) and each eigenvalue of $A$ is equal to $\pm 1$. Why is $A$ similar to $A^{-1}$? (by Jordan form)

Martin Sleziak
  • 53,687
user241906
  • 21
  • Double post. cf. http://math.stackexchange.com/questions/1284461/if-each-eigenvalueof-a-is-either-1-or-1-rightarrow-a-is-similar-to/1284551#1284551 –  May 19 '15 at 19:01

2 Answers2

2

Note that if $$ A = \pmatrix{A_1\\&\ddots\\&&A_n} $$ then $$ A^{-1} = \pmatrix{A_1^{-1}\\&\ddots\\&&A_n^{-1}} $$ So, it suffices to show that a Jordan block on $1$ or $-1$ is similar to its own inverse.

For now, take $$ J = \pmatrix{1&1\\&1&1&\\&&&\ddots &1\\ &&&&1} $$ Following the explanation here, we find $$ J^{-1} = \pmatrix{ 1&-1&1&-1&\cdots\\ &1&-1&1 & \cdots\\ && \ddots\\ &&&&1 } $$ It then suffices to note that $J^{-1}$ has $1$ as its only eigenvalue and that the matrix $J^{-1} - I$ has rank $n-1$.

A similar argument can be applied to the Jordan block on $-1$.

Community
  • 1
Ben Grossmann
  • 225,327
0

Suppose $Ax=\lambda x$. Then $A^{-1} (Ax) = x = A^{-1} \lambda x = \lambda A^{-1} x.$ So $1/\lambda$ is an eigenvalue of $A^{-1}$ with the same eigenvector $x$. But if $\lambda = 1$ then $1/\lambda = 1$, or if $\lambda = -1$ then $1/\lambda = -1$. So $A^{-1}$ has the same eigenvalues as $A$ with the same geometric multiplicities. Can we conclude that they are similar from here? (We certainly can in the diagonalizable case.)

Ian
  • 101,645