I've been doing a bit of work these past couple of nights on computing cyclic subspace decompositions, finding cyclic bases, and then computing the Jordan canonical form of matrices.
My question is: When I compute, say, generalized eigenvectors that are in the nullspace of $(A-\lambda I)^2$, call this nullspace $K_2$, and I pick a vector from $K_2$ that is not in $K_1$, which is the null space of $(A-\lambda I)$, why is it called, "the vector is in $K_2$ mod $K_1$".
I've learned the word "modulo" in an introductory proofs class as something like this: $x$ is congruent to $y$ mod $z$, if $(x-y)/z$ yields an integer.
But I don't see how this definition helps with my understanding of "$K_2 mod K_1$".
Also, I think that $K_2$ mod $K_1$ is a "quotient space", which vaguely means that $K_1$ collapses to the zero vector - so does this mean that $K_2$ mod $K_1$ is just a "smaller" vector space, the vector space $K_2$ + a zero vector, $K_1$?
Thanks,