Linear Algebra 7 | Examples for Subspaces

Рет қаралды 34,765

Күн бұрын

Пікірлер: 29

@MrWater2 Жыл бұрын

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!

@johnartzi5693 2 жыл бұрын

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

@jacktrainer4387 2 жыл бұрын

Thanks!

@brightsideofmaths 2 жыл бұрын

Thank you very much :)

@nightmarechameleon7502 Жыл бұрын

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

@brightsideofmaths Жыл бұрын

Nice :) And thank you for the support! :)

@Leslie.Green_CEng_MIEE 2 ай бұрын

At 5:41 "Is U closed under vector addition?" the label reads "(b)" but should read "(c)" since we already had a (b), and the (c) refers back to Linear Algebra 6, "Characterisation for subspaces" section. This is a small point, but may help with the notes. There also seems to be a 1 in 30,000 chance that anyone would mention it! An excellent set of videos, ... as usual 😊

@brightsideofmaths 2 ай бұрын

Wow, thanks a lot. I really didn't see this typo before :) I will correct it in the pdf versions.

@林志雄-v1u 7 ай бұрын

謝謝林益辰教授讓我認識自主學習的重要❤❤❤

@khalidmohammadjama 2 жыл бұрын

Thank you

@narendraparmar1631 10 ай бұрын

Thanks for the efforts

@brightsideofmaths 10 ай бұрын

My pleasure

@darcash1738 10 ай бұрын

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?

@DutchMathematician 9 ай бұрын

I don't know what exactly you mean with λ(λx1 = x2) This is not a mathematically correct expression. Let me show you how I would prove that U is not a (linear) subspace in a rigorous way (without giving a counter example). Let x=(u,v) be an arbitrary element of U. This implies that v=u². If λx were an element of U, then (by definition of U) we should have λv = (λu)² ⟺ λv = λ²u² Now use the fact that x is an element of U, hence that v=u². We then get: λv = λ²v Subtracting λv from both sides and factoring out the common factor λ we get: λ(λ-1)v = 0 (*) Note that x=(u,v) was an arbitrary element of U. We could have chosen x=(1,1) (obviously an element of U). Hence, λ must be either 0 or 1 for (*) to hold in general. In particular, it does not hold for λ=2 for a general vector x. Hence, U is not a subspace. The proof above shows that U is not a subspace. However, it is (in general) way more easier to prove that a set U is not a subspace by giving just one counter example: one such an example is sufficient! (even proving that 0 is not an element of U suffices (although for the U given in the video, 0 actually *is* an element of U)). Of course, you could also take two arbitrary vectors x and y in U and show that x+y is not an element of U. Now your last question: "what are some other good counterexamples and structures like that which will always fail?" Well, any subset U that is defined by a restriction given by a general function f(u,v)=0 where f is *not* a linear function will not be a subspace. I hope this was helpful.

@darcash1738 9 ай бұрын

@@DutchMathematician⁠⁠ thank you. Also for this strategy you have we just have an input and output in the form u, v. Would it ever be more than that? For a differential equation for example would the input just be dy/dx (or whatever var) and the output the general solution? Also I still don’t get why that way I did doesn’t work mathematically, I just factored out the scalar lambda. I believe that in itself would work mathematically but it was probably an earlier step that is actually wrong. Because you also factored out lambda in your way too. I plugged the lambda into the given equation of x1^2 = x2. Could you tell me why this doesn’t work or if it was a step in between that and factoring out the lambda

@DutchMathematician 9 ай бұрын

@@darcash1738 I don't understand what you mean with 'input and output', when talking about a vector x in the two dimensional real space with entries u and v. There is no function involved here. We just want to show that λx in general is not an element of U for arbitrary λ and x. (let alone talking about a derivative...) With respect to your second remark... As I mentioned earlier, "λ(λx1 = x2)" is *not* a mathematically correct expression: it just doesn't make sense. The expression "λ(λ-1)v = 0" on the other hand *is* a mathematically correct expression and *does* make sense.

@darcash1738 9 ай бұрын

@@DutchMathematician idk what you mean by the function f(u, v) then. What are the restrictions on this function?

@DutchMathematician 9 ай бұрын

@@darcash1738 For the subset U from the video we can take the function f to be f(u,v)=u²-v. Then the equation f(u,v)=0 restricts the components u and v from an arbitrary vector x=(u,v) exactly as the subset U is defined.

@GeoffryGifari 2 жыл бұрын

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?

@nathanwycoff4627 2 жыл бұрын

regarding your final question, notice that we can rewrite (x1,x2,x3) as u(1,1,1) + v(0,-1,1). This shows us that we indeed have a linear subspace :)

@brightsideofmaths 2 жыл бұрын

We will talk about free variables a lot in this series :)