Given a basis B={v1,..,vn} for R^n. Prove that for every basis C in R^n the change of coordinate matrix M from basis B to C is invertible. my proof: Let C={c1,…,cn} be a basis for R^n ,and T:R^n→R^n an operator such that Tx=Mx where M is the change of coordinate matrix from basis B to C. (i)M is built from n linearly independent vectors so the rank(M)=dim(R^n )=n therefor T is onto (ii)The dimension of domain and the range of T is n ,and from (i) we can conclude that T is one-to-one. Form (i) and (ii) T is bijective so its an isomorphism there for T is invertible so the representation matrix M is invertible.
can someone please tell me if this proof is acceptable,and if not how can i make it better?
Jace Butler
Please prove this isn't your homework.
>Protip: you cant
>Protip: chegg you fucking jew
>sage
Leo White
this is a homework yes ,i just wanna hear thoughts on my solution kind user.
Alexander Myers
no idea, but i for fun proved it myself:
let T be the transformation matrix from B to C.
Now it goes:
T not invertible dim Kernel (T) > 0 T x = 0, x != 0 T (sum a_i b_i) = 0, with a_i in R and b_i as basis vectors of B = {b_1, b_2, ..., b_n}
T is linear therefore we conclude that T (sum a_i b_i) = sum a_i T(b_i) = sum a_i c_i = 0, with basis vectors c_i of basis C. the latter equation means that the basis is not linear independent in contradiction with the assumption that it is a basis of C.
therefore every transformation matrix is invertible.
Christopher Brooks
Always feels so good to see how others approach proofs ,thank you for sharing user!
Bentley Miller
>Always feels so good to see how others approach proofs ,thank you for sharing user! your welcome, i love doing proofs every once in a while despite SUCKING HARD (im not being polite, literally everyone seems to be better than me at it) at it :)
Joseph Baker
This. Shit shouldn't warrant a reply.
Cameron Watson
This would be trivial if you had a linear algebra clasa that constructed things properly because a transformation is invertible if and only if it's asociated matrix is invertible. And the identity transformation is pretty invertible.
Hudson Wilson
its surjective. qed.
Why are you writing so much for such an obvious thing?
Anthony Price
We know that B is a matrix of full rank, and that C must be of full rank as well to constitute bases for R^n. Compute the eigenvalue decomposition of M=U*D*U^-1. We know that hence det(M*B)=det(D)*det(B)=det(C). C must have a nonzero determinant since it must be of full rank (all columns are orthogonal). Suppose M is not of full rank. Then, det(D) and therefore det(C) are 0 since det(B) is nonzero, which would imply that C is not a basis for R^n. Hence, det(D) must be nonzero which means that M must be of full rank.
Chase Mitchell
why are you writing so much? what does it have to do with eigenvalues?
The proof is literally a one-liner.
Josiah Sullivan
Only the zerovector is a solution to Mx = 0, thus it is invertible. QED. ?? Pretty simple.
Owen Davis
Homework threads warrant a ban OP.
Luis Gray
But fucking anything can be homework you idiot.
Angel Lopez
For any two bases there is a change of coordinate matrix that changes from one to the other. Invertibility literally follows from that.
Grayson Barnes
I'm surprised that you guys haven't done something like: For finite dimensional operators, injectivity=surjectivity=bijectivity=invertibility. So all you have to do is prove one to get invertibility.
Jace King
I think what you're trying to imply with your post is that all transformations from one basis to another are the identity transformation, which is incorrect.
Nathan Parker
not him but what shows M is linearly independent?
Samuel Campbell
If there exists a matrix M such that Tx=Mx where T is a linear transformation, then M must be constructed of linearly independent column vectors, which implies that M is invertible by the fundamental theorem of invertible matrices....isn't that all you need? someone correct me
Hudson Martinez
[math](id)_{CB} \circ (id)_{BC} = (id)_{BC} \circ (id)_{CB}= id_{\mathbb{R^n}}[/math] so [math](id)_{CB}[/math] is an isomorphism. qed.
Jason Jones
No. Only the zerovector is a solution. Because: Each of the basis are linearly independent so for the zerovector, only the zerovector are it's coordinate vector in each of the basis. There can be no vectors that has the zerovectors as a coord-vector for the B-basis that is not itself the zerovector, likewise there can no vector that has the zerovector as a coord-vector for the C basis that is not itself the zerovector. And because all vectors can be written as a coord-vector wrt B and C, and all coord-vectors in B or C map to one vector in our vector space, there can only be one vector that gets the zerovector as the coordinate vector in basis C, and there can be only one coordinate vector in B that gives you the zerovector in our vectorspace. Thus we can see that the zerovector is the only solution to Mx = 0, and thus by the IMT M is invertible.
Daniel Howard
If Mx = 0 has a nonzero solution then there is a nonzero vector x1 in R^n which maps to x as its coordinate vector wrt basis B. That is x is the B-coordinate vector of x1, and x1 is nonzero.
x1 is nonzero because if it was zero then x would be zero too as x = C^(-1)x1. where C is the matrix with columns from the c-basis.
However as Mx gives the C-coordinate matrix of x1 then, the C-coordinate matrix of x1 is 0. However since we can find x1 from the c-coordinate matrix of x1 by a maxtrix-vector product (which one?) and the c-coordinate matrix of x1 is 0, x1 must also be zero.
Thus we have a contradiction, and so Mx = 0 can have only the trivial solution. Thus by the IMT M is invertible.