/sci/ - Science & Math

>>8677825
>i lost my ability to abstract and create. just trying to get it back

you started with the yoneda lemma?
i admire your boldness
but mathematics is like mountain climbing
if you don't start at the bottom and acclimate on your way up, you're going to suffocate and freeze to death

if you don't know category theory, you obviously won't understand the yoneda lemma
do not be dissuaded, algebra and category theory is a fun journey
but you must start at the beginning, wherever that is for you

if you're good with algebra, you could probably jump right in to some of the more advanced books like categories for the working mathematician

algebra: chapter 0 is a good book all around, but the exercises can be hard for a true beginner

>>8240923
given an [math] \mathbf{A} \in \text{GL}_n(\mathbb{C})[/math], the vector space [math] V [/math] of [math] n \times n [/math] matrices over [math] \mathbb{C} [/math], and a linear map [math] F:V\to V~,~ F\mathbf{M = A^{-1}MA} [/math]

first, look at the determinant of [math] F\mathbf{M} [/math]:
[math] F\mathbf{M = A^{-1}MA} \implies \det(F\mathbf{M})=\det(\mathbf{A^{-1}MA}) [/math]
[math] \implies \det(F)\det(\mathbf{M}) = \det(\mathbf{M}) \implies \det(F) = 1 [/math]

now, we're looking for eigenvalues of F, which in this case are complex numbers [math] \mathbf{z} [/math] such that [math] F\mathbf{M = zM} [/math]

take the determinant of both sides [math] \det(F\mathbf{M}) = \det(\mathbf{zM}) [/math]
recall that [math] \det(F\mathbf{M}) = \det(\mathbf{M}) [/math]
so we have [math] \mathbf{\det(M)=\det(zM)} [/math]
by a determinant property, [math] \det(\mathbf{zM}) = \mathbf{z}^n\det(\mathbf{M}) [/math]
[math] \det(\mathbf{M}) = \mathbf{z}^n\det(\mathbf{M}) \implies \mathbf{z}^n=1 [/math]
so the solutions to this, and thus the eigenvalues of F, are the roots of unity

is that correct?