So what's the point of linear algebra, anyway?
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Re: So what's the point of linear algebra, anyway?
#2People have written very nice books on this e.g.
Re: So what's the point of linear algebra, anyway?
#3People have written very nice books on this e.g. https://linear.axler.net/
> The novel approach taken here banishes determinants to the end of the book.
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?
Re: So what's the point of linear algebra, anyway?
#4People have written very nice books on this e.g. https://linear.axler.net/
> The novel approach taken here banishes determinants to the end of the book. 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? [0] https://www.youtube.com/watch?v=Ip…
One does not need determinant to define eigenvalues. For example:
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.
Re: So what's the point of linear algebra, anyway?
#5Earlier quoted context omitted.
> The novel approach taken here banishes determinants to the end of the book. 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? [0] https://www.youtube.com/watch?v=Ip…
One does not need determinant to define eigenvalues. For example: 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.
I agree, but the definition alone isn't sufficient to actually calculate eigenvalues. Hence the standard approach which says that for matrix A, vector v, and eigenvalue λ, we have
Av = λv
=> Av - λv = 0
=> (A - λI)v = 0
=> det(A - λI) = 0
Re: So what's the point of linear algebra, anyway?
#6Earlier quoted context omitted.
One does not need determinant to define eigenvalues. For example: 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.
I agree, but the definition alone isn't sufficient to actually calculate eigenvalues. Hence the standard approach which says that for matrix A, vector v, and eigenvalue λ, we have Av = λv => Av - λv = 0 => (A - λI)v = 0 => det(A - λI) = 0 Which then yields the characteristic polynomial. Skipping the determinant means you need a different approach.
You can prove many fundamental results of Linear Algebra with the definition that does not directly use determinants. In fact, one would define trace and determinant as sum and product of eigenvalues. Definition of characteristic polynomial would then follow.
If "computation" is what you are after then Av = λv is about solving a system of equations and you can try elimination, etc.