Navigating Linear Algebra Ep. 3: Linear Maps

[@LetsSolveMathProblems]

The problems for this video (please try out at least one!) can be found in this link: github.com/LetsSolveMathProblems/Navigating-Linear-Algebra/blob/main/Episode%203.pdf. Again, I encourage you to post solutions to these problems in the comment section, as well as peer-review other people's proofs. :)

[@ummwho8279]

I haven't watched the video yet and will tomorrow, however, I'm going to take a try at Problem 3 since it looks easy (although looks can be deceiving!!) based on what I already know and will compare tomorrow to see if I missed anything after watching the video. Problem 3: a) is ii) reflection about the line y=x b) is i) reflection about the x-axis c) is iii) rotation by 180 about the origin.

[@cantcommute]

+1

[@LetsSolveMathProblems]

For Problem 3, this is correct. :)

[@cantcommute]

For problem 2: We need to show that for any scalar λ in the field, ϕi(λv+w)=λϕi(v)+ϕi(w). Where ϕi is the inverse of ϕ and v,w in W. (We can always set w=0 or λ=1.) To show that this works just note that because ϕ is isomorphic we can write v=ϕ(a) and w=ϕ(b) for a,b in V. Hence, λv+w=λϕ(a)+ϕ(b)=ϕ(λa)+ϕ(b)=ϕ(λa+b). So ϕi(λv+w)=ϕi(ϕ(λa+b))=λa+b=λϕi(v)+ϕi(w) as desired.

[@LetsSolveMathProblems]

For Problem 2, this is correct and well-written! :)

[@fierydino9402]

Your lectures are marvelous!!! I wish I could learn various subjects from you! Do you have any plans to do the lecture series on topology or group theory?

[@LetsSolveMathProblems]

Thank you for the compliment! I actually have an in-progress series on metric space topology, so I'll link it here in case you are interested: kzbin.info/aero/PLpoKXj-PWCbZHDDbUuZCNoocz5k-ovqy2. In the future, I absolutely have plans to make lectures on group theory and more advanced branches of topology, although my uploading schedule has been admittedly a bit slow so far. :)

[@walkingradiance9556]

Does the isomorphism form relate to the groebner basis?

[@cantcommute]

For problem 4: Hom(V,F) is defined as the set of functions that take vectors from the vector space V to the field F. Function addition is just adding functions together while keeping the arguments the same. Then, it is easy to see that this will form a vector space; as every function f in Hom(V,F) is in F itself given an argument. So it will satisfy all addition, scalar and distributive properties. (Caveat: just note that the zero vector is just a function that takes everything to 0 which is obviously linear. While the existence of the negative vector is just multiplying by -1, as that must also be in F and is a valid linear function from V to F.) Update (see first comment): Forgot to note that if f,g are in Hom(V,F) then the linear combination af+g will also be in there for a in F. To see this just note that af is obviously linear. Also for v,w in V; h(v+w)=af(v+w)+g(v+w)=af(v) + g(v) + af(w) + g(w) =h(v) + h(w) thus it's in Hom(V,F).

[@LetsSolveMathProblems]

Yup, but you likely also want to add that if f and g are in Hom(V, F) and a is real, then af + g is linear.

[@cantcommute]

@@LetsSolveMathProblems ohhh yeah I knew I was forgetting something. That was actually the first thing I thought of but then managed to un-think it lol

[@cantcommute]

For problem 1: If ϕ is injective and ψ is surjective then ϕ(ψ(x)) is not necessarily injective as we may have two values for x for which ψ(x) has the same value and so ϕ∘ψ(x) can be not injective. ϕ might also not be surjective so it will always miss some values and thus ϕ∘ψ(x) can be neither surjective nor injective. (Update: it would be nice if somebody can post an example!) But, if they were linear: As it turns out (had to think about this a lot) the only linear function from R to R is the linear line about 0. To see this: f(x)=f(x*1)=xf(1) and that f(0)=0. If f is a function that is either surjective or injective then f(1)!=0. Hence f(x) is just bijective. The composition of two bijective functions is obviously bijective so ϕ∘ψ can't be neither surjective nor injective.

[@LetsSolveMathProblems]

Both parts are correct, with one small caveat: Technically, you have to show existence of a function f: R \to R that is surjective yet not injective (and vice versa), but this isn't too hard.

[@SphereofTime]

7:45

[@axemenace6637]

In problem 4, the question describes linear maps from V to F. However, a linear map must be between one vector space and another vector space, as I understand it. But F isn't really a vector space - it is a field of scalars. So, it seems to me that the only way for this to make sense is to consider the vector space from F over F, in which case the scalars are also the vectors, which seems somewhat nonsensical. If V is a vector space over F, how can I understand linear maps from V to F?

[@LetsSolveMathProblems]

Not quite: F is indeed a vector space over F! Just let the vector space operations be defined to be field addition and field multiplication. It might've been the best if I had specified it; apologies for any confusion.

[@plusjeremy]

Apologies if you have talked about this before: can you share the equipment and software you use to make these videos? Thanks!

[@LetsSolveMathProblems]

At the moment, I'm using Wacom Tablet, Krita (for whiteboard), Camtasia, and ~$30 USB microphone.