Can Any Matrix Be Similar to Itself Without Being a Scalar Multiple?

[td21]

If $S^{-1}BS=A$, can we always find Q ,such that $Q^{-1}BQ=A$? Q different form S and no scalar multiple of S.
If not, what are some special cases in which we can find one?
i can only find one when A is nonidenityly similar to itself.

 

[henry_m]

This is equivalent to the question of when we can find a similarity transformation that leaves a matrix A invariant: P such that $P^{-1}A P=A$, with P not proportional to the identity ($P=Q^{-1}S$). For complex matrices, the easiest way to answer the question is to put the matrix in Jordan normal form. The similarity transformations P are then a combination of:

i) Transformations that exchange the order of the Jordan blocks, and
ii) Self-similarity transformations within a single Jordan block.

The second type consists of all upper triangular matrices with entries constant on each diagonal, and nonzero on the main diagonal (so it's invertible). This creates quite a big family of transformations.

[This is often very relevant in the representation theory of groups and other algebraic structures when asking when there are P's which do this for many matrices simultaneously; look up Schur's lemma.]