Linear Algebra 7 | Examples for Subspaces 

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

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 :)

My pleasure

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.

Thank you very much :)

Its a base concept which unites a few axioms that are very helpful for countless things