Linear Algebra 7 | Examples for Subspaces
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- เผยแพร่เมื่อ 16 พ.ค. 2022
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(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)
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!
This feels much easier to understand then your other videos I like it! Keep the series going
Thank you
I've been stuck on this topic for a bit, and I have to say, the thoroughness presented here really helped! Kudos!
Nice :) And thank you for the support! :)
謝謝林益辰教授讓我認識自主學習的重要❤❤❤
Awesome 😮
I love it! :o
Thanks for the efforts
My pleasure
Awesome!!! :-)
Thanks!
Thank you very much :)
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?
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 :)
We will talk about free variables a lot in this series :)
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?
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.
@@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
@@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.
@@DutchMathematician idk what you mean by the function f(u, v) then. What are the restrictions on this function?
@@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.
Why study subspace. Don't think it has any significance and application.
Its a base concept which unites a few axioms that are very helpful for countless things