Navigating Linear Algebra Ep. 4: Basis

[@LetsSolveMathProblems]

The problems for this video can be found in this link: github.com/LetsSolveMathProblems/Navigating-Linear-Algebra/blob/main/Episode%204.pdf. Again, I encourage you to post solutions to these problems in the comment section, as well as peer-review other people's proofs. :)

[@fierydino9402]

I really enjoy your classes! They are so clever! Thank you for offering them.

[@mattb9557]

Awesome

[@yassinezaoui4555]

Pb 5: I think it's yes, there exists. For exemple, let A= (0 1 0 ) (0 0 1 ) (0 0 0 ) we can easily calculate A²= (0 0 1) (0 0 0) (0 0 0) and A^3 = 0 zero mattrix of M3(R) We can see that kerA= Vect[e1] , KerA² = vect[e1,e2] and KerA^3 = vect[e1,e2,e3] = the whole space, so we have indeed kerA kerA² and kerA²kerA^3

[@LetsSolveMathProblems]

For Problem 5, this is correct.

[@yassinezaoui4555]

@@LetsSolveMathProblems thx ^^

[@cantcommute]

For problem 2: Just notice that if you have an isomorphism between two vector spaces V and W using ϕ and v1,v2,...,vn as the basis for V then ϕ(v1),ϕ(v2),...,ϕ(vn) is a basis for W. Hence they would have to have the same dimension. To prove this: for scalars a1,a2,...,an in F; every v in V can be written as v=a1v1+a2v2+...+anvn. Notice that ϕ(v)=a1ϕ(v1)+a2ϕ(v2)+...+anϕ(vn). If v=0 then due to linear independence a1=a2=...=an=0. By definition ϕ(0)=0 and is the only vector that does that due to being an isomorphism. Thus a1ϕ(v1)+a2ϕ(v2)+...+anϕ(vn)=0 implies a1=a2=...=an=0. Due to ϕ being an isomorphism there is a unique mapping to each vector in W. So ϕ(v1),ϕ(v2),...,ϕ(vn) also spans W as we can get to any vector in W using a vector V using the isomorphic mapping of ϕ. Hence it is a basis. Hence it has the same dimension n.

[@LetsSolveMathProblems]

For problem 2, this is correct.

[@cantcommute]

For problem 4: To construct a guess, take a function f in this dual space. Applying it to the vector [x,y] then gives back f([x,y])=xf([1,0])+yf([0,1]). Hence it's just the set of planes in 3-space! To get all possible functions, then, you can use the basis of the functions g,h such that g([1,0])=1,g([0,1])=0 and h([1,0])=0,h([0,1])=1. They will both be linearly independent, and a linear combination will give back all possible planes.

[@LetsSolveMathProblems]

For problem 4, this is correct.

[@cantcommute]

For problem 3: Kinda confused about this one. I think the problem probably has to do with not being able to shrink vectors. In this construction [2,3] and [4,6] would be linearly independent from each other because you cannot get to [2,3] from [4,6] (you can the other way though?). But -2*[2,3]+[4,6]=0. Update (see first comment): You can't get from [4,6] to [6,9] and vice versa but 3*[4,6]-2*[6,9]=0.

[@LetsSolveMathProblems]

Not quite! As you said, you can go "the other way" from [2, 3] to [4, 6]. Can you change the two vectors such that they're still linearly dependent by standard definition, but you can't get one from the other and vice versa?

[@cantcommute]

@@LetsSolveMathProblems oh.

[@LetsSolveMathProblems]

@@cantcommute Your updated solution is correct. :)

[@cantcommute]

yaaaas

[@cantcommute]

For problem 1: It should be infinite-dimensional because you cannot get to different real numbers whose ratios aren't in Q (true also when you have a linear combination, though this is hard to say and it's what we're trying to prove.) There is an infinite number of those. Suppose it's finite-dimensional with dimension n. Then one may choose an n vector basis (of real numbers) and make any other vector (real number) as a linear combination using the field (rational numbers). For a given choice of scalar constants (rational numbers), notice that this is a unique vector (real number). To see this: suppose it is not unique, then we can write w=a1v1+a2v2+...+anvn=b1v1+b2v2+...+bnvn iff (a1-b1)v1+(a2-b2)v2+...+(an-bn)vn=0. By definition of a linear independence ak=bk for all integer k. Hence it is unique. This then shows that there exists an injective map to R from Qn. Notice that it must also be surjective as it spans R. Hence it is bijectve. This is not allowed, as Qn is bijective to Nn which is bijective to N which is not bijective to R. Succinctly: they have different cardinalities.

[@LetsSolveMathProblems]

For problem 1, this is correct.

[@yassinezaoui4555]

Pb2: We want to prove that if V and W are finite-dimensional and have different dimensions, then there cannot be an isomorphism from V to W. This equivilent to prove that if there is an isomorphism from V to W which are finite dimensional, then dimV = dimW. So, let's suppose f is an isomorphism from V to W and dimV = n. dimV=n so we can consider B=(e1,...,en) a basis of V and let's try to prove that C=(f(e1),...,f(en)) is a basis of W which leads to dimW = n dimV. LI of C: Let a1, ..., an be some scalars such that a1.f(e1)+...+an.f(en) = 0 (zero vector of W) so since f is linear we can write f(a1.e1+....+an.en) = 0 = f(0) and since f is injective ( it is an isomorphism) then a1.e1+...+an.en = 0 (zero vector of V) but since B is a basis of V then a1=...=an=0 . Hence, C is LI. C spans W: Let w be a vector of W, since f is surjective (f is an isomorphism) there exists a v in V such that f(v)=w. B is a basis of V then there exists scalars b1,...,bn such that v = b1.e1+...+bn.en. w = f(v) = f(b1.e1+....+bn.en) = b1.f(e1)+...+bn.f(en). so for any vector in W we can write it as a linear combination of f(e1),....,f(en) thus C spans W. C is LI and spans W => C is a basis of W =>dimW = card(C) = n = dimV

[@LetsSolveMathProblems]

For Problem 2, this is correct! :)

[@yassinezaoui4555]

​@@LetsSolveMathProblems thx ^^

[@cantcommute]

Nice I had the same method/answer

[@yassinezaoui4555]

Pb4: Let's try to show that B=(a,b) is a basis of (R²)* where a: (x,y) in R² ---> x in R and b: (x,y) in R² ----> y in R. LI of B: let n and m be reals such that n.a + m.b = 0 zero vector of (R²)* meaning 0: (x,y) in R² ----> 0. so for every (x,y) in R², n.a(x,y)+m.b(x,y)=0 =>n.x+m.y=0 , since this is true for every reals x and y we can seek good choices to end up with n=m=0 for exemple we can take first x=y=1 and second x=1 and y=-1 so we have this simple system n+m=0 and n-m=0 , adding the two equations we end up with 2.n=0 => n = 0 and from the 1st eqution m=-n=0 so n=m=0 => B is LI B spans (R²)*: let c be a linear map in (R²)*. then c: (x,y) in R² ---> c(x,y) in R c(x,y) = x.c(1,0) + y.c(0,1) = r.a(x,y) + s.b(x,y) where r=c(1,0) which is a real number and s=c(0,1) which is a real number as well, so for every (x,y) in R² we can write c(x,y) as a linear combination of a(x,y) and b(x,y) so we can write c as a linear combination of a and b as a result B spans (R²)*. B is LI and spans (R²)* => B is a basis of (R²)*

[@LetsSolveMathProblems]

For Problem 4, this is correct.

[@yassinezaoui4555]

​ @LetsSolveMathProblems thx ^^

[@cantcommute]

For problem 5: The following matrix works: |0 1 0| |0 0 0| |1 0 0|

[@LetsSolveMathProblems]

For problem 5, this is correct.

[@axemenace6637]

I'd like to point out that this video does not establish that the dimension of a vector space is well-defined. In particular, you do not explicitly establish that 1. there exists a finite-dimensional linearly independent set which also spans V, and you do not show that 2. every such set is of equal size. I would invite all the viewers to prove these claims. Note that the first claim is only (necessarily) true if there exists a spanning set of finite size. Otherwise, the existence of a basis is equivalent to the Axiom of Choice, and thus cannot be explicitly proven.

[@LetsSolveMathProblems]

This is true, and I thank you for pointing this out. I hope that viewers who take time to parse this episode and try out the problems can show many of these claims (e.g., Problem 2 is very much related to your second point), other than those for infinite-dimensional vector spaces, of course, for which you indeed need the axiom of choice.