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3PS | 1 year ago
Big fan of this approach! Though I have warmed up to determinants ever since I saw 3Blue1Brown give a fairly intuitive explanation for them [0].
I'm kind of curious as to how they covered eigenvalues/the characteristic polynomial without determinants. Maybe they just jumped straight to diagonalization?
richrichie|1 year ago
If T is a linear operator on vector space V, a scalar a is an eigenvalue if there is a v in V s.t. Tv = av.
This is the approach the book takes.
3PS|1 year ago