Undergrad How to prove that shear transform is similarity-invariant?

[swampwiz]

OK, my understanding is that the characteristic equation for a matrix is similarity-invariant, from which results that the trace & determinant - which correspond to the penultimate & free coefficients of the characteristic equation - and other parameters (corresponding to other coefficients) are invariant under a similarity product of any matrix, And since any matrix can be composed of a concatenation of swap, scale or shear transforms, it must be that the characteristic equation is invariant under a similarity-transform (i.e., so that all of its coefficients are invariant).

I have worked out that the swap & scale similarity products result in the same characteristic equation, but am having difficult doing so for the shear. As part of the expression of the characteristic equation of the similarity product I get the characteristic equation of the original (i.e., central) matrix, but I get a bunch of other terms that should resolve to zero, but I just don't see it, at least from my math.

I've been searching for a good source online that goes into proving this, but all I see is proofs that the determinant is invariant, but not the characteristic equation. The proof of the determinant is simple; I don't need to see that. But simply changing the matrix to the characteristic matrix by adding in the characteristic parameter (i.e., eigenvalue parameter) seems to make this much more difficult.

Thanks

 

[Erland]

The characteristic polynomial $\det(\lambda I -A)$ , where $A$ and $I$ are $n\times n$-matrices, is invariant w.r.t. similarity transformations, which means that if $P$ is any invertible $n\times n$-matrix, then $\det(\lambda I-P^{-1}AP)\equiv\det(\lambda I-A)$. You say you are familiar with the corresponding proof for determinants. Can you adjust that proof to find the proof for the characteristic polynomial? I see no reason to look separately at shear-transformations, swaps, scale transformations, etc.

 

[swampwiz]

I am basing the proof of the determinant of just the matrix on the fact that the determinant of a matrix having a repeated row is 0. I don't think that I can say that the determinant of the characteristic matrix of such an original matrix that has its rows swapped is 0. However, I can't say that the determinant of that characteristic matrix is always zero unless I can remove the characteristic parameter completely, which I can't seem to be able to do.

 

[Erland]

There is a much simpler way. You already know that $\det(P^{-1}AP)=\det(A)$. Apply this with $\lambda I-A$ instead of $A$, and rewrite it a little.

 

[swampwiz]

There is a much simpler way. You already know that $\det(P^{-1}AP)=\det(A)$. Apply this with $\lambda I-A$ instead of $A$, and rewrite it a little.

Thanks. I knew I was making it a lot harder than I needed to.

 

[mathwonk]

Don't feel alone. I am a professional mathematician but as a student I also did not quickly react to things like noticing that all matrices commute with a matrix like cI, leading to the cancellation that makes this little thing pop out. I.e. it was not at all obvious to me that P^-1(cI-A)P = (cI - P^-1AP).