## Linear Algebra Change Matrices Confusion

1. The problem statement, all variables and given/known data
Suppose N is an invertible n x n matrix, and let D = {f1, f2, ... , fn} where fi is column i of N for each i. If B is the standard basis of Rn, show that MBD(1Rn) = N.

Call the standard basis of Rn = {E1, ... , En}

2. Relevant equations

3. The attempt at a solution
The first thing I don't get is whether D is a basis. I thought it had to be a basis to do this kind of question, but the problem doesn't specify! I'm going to assume it is...

Now I'm going to write the matrix M, specifying its entries. For example, f11 is the entry at row 1, column 1. f1 will just denote column 1 of M.
M =
[f11 ... f1n
: :
: :
fn1 ... fnn]

1Rn(f1) = f1, 1Rn(fn) = fn.

f1 can be written as f11E1 + ... + fn1En
fn can be written as fn1E1 + ... + fnnEn

Then MBD(1Rn) is the coefficients of the above, written in column form, so we get exactly the matrix M.

This seems to prove it! But the question specifies that M is invertible, and I didn't use that fact at all. So I think I may have done something wrong. Can anyone help?

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 Recognitions: Homework Help Science Advisor N is invertible if and only if the columns are a linearly independent set. You didn't use that N is invertible because you just assumed D is a basis.
 Ooh, right - I completely forgot about that. I'm confused about what D is actually a basis FOR. Is it a basis of R^n? If so I can say that because there are n elements in D, and it's linearly independent, it spans. Therefore it's a basis. If it's not a basis of R^n, I'm not sure.

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## Linear Algebra Change Matrices Confusion

 Quote by jumbogala Ooh, right - I completely forgot about that. I'm confused about what D is actually a basis FOR. Is it a basis of R^n? If so I can say that because there are n elements in D, and it's linearly independent, it spans. Therefore it's a basis. If it's not a basis of R^n, I'm not sure.
Well, sure. If D is invertible its columns have to span R^n. Otherwise it wouldn't be onto, would it?

 What does onto have to do with it? I thought it had to span because A is invertible if and only if AX = B has a solution for every B, not anything to do with transformations...

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