Linear Algebra 7 | Examples for Subspaces

Thanks to all supporters! They are mentioned in the credits of the video :)

This is my video series about Linear Algebra. I hope that it will help everyone who wants to learn about it.

#LinearAlgebra
#Vectors
#Matrices
#MachineLearning
#Eigenvalues
#Calculus
#Mathematics
x

(This explanation fits to lectures for students in their first year of study: Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)

[MrWater]

What a good example and excellent explanation. I was doing the proof by myself and then I watch the proof and note that I didn't understand properly what and specially how I have to do it. I really like how you explained us how to do something, that's incredible helpful. I hope you never stop of doing this videos (well at least not in the near future!) it's amazing how much I'm learning thanks to you!

[johnartzi]

This feels much easier to understand then your other videos I like it! Keep the series going

[nightmarechameleon]

I've been stuck on this topic for a bit, and I have to say, the thoroughness presented here really helped! Kudos!

[darcash]

Nice and would we be able to prove that last bit in general like this or did I mess up any steps?:
(λx1)^2 = λx2 [Replace x1 with λx1 and x2 with λx2]
λ^2*x1^2 = λx2 [Distribute the square]
λ(λx1 = x2) [Factor out λ]
For λ ≠ (0, 1), it fails, showing it is not a subspace.

Also, what are some other good counterexamples and structures like that which will always fail?

[GeoffryGifari]

hmmm i see a pattern... in the subspace example the vector in Rⁿ seems to only need only one free variable in R for one component while every other component can be built through (nonzero) scalar multiplication of that one variable (maybe this is a special case)

if for example we have the condition of fixing one component of the vector to be a nonzero constant (like x₃ = 5) we already violate the first and second criteria for a subspace. having one component to be a nonlinear function of another also violates the conditions as shown.

what if we make something like this: (x₁, x₂, x₃) is an element of R³ with the condition x₁ = u, x₂ = u-v, x₃ = u+v for every u, v in R?

[rbc]

Why study subspace. Don't think it has any significance and application.