- #1
swampwiz
- 571
- 83
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
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