# Linear Algebra Change Matrices Confusion

1. Apr 12, 2010

### jumbogala

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?

2. Apr 12, 2010

### Dick

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.

3. Apr 12, 2010

### 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.

4. Apr 12, 2010

### Dick

Well, sure. If D is invertible its columns have to span R^n. Otherwise it wouldn't be onto, would it?

5. Apr 12, 2010

### jumbogala

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...

6. Apr 12, 2010

### Dick

Sorry, I mean N is invertible only if its columns D span R^n. N(E1)=f1. N(E2)=f2. Etc. So N(a1*E1+...+an*En)=a1*f1+...+an*fn. If the f's don't span, then there is a B that's not a combination of the f's. Then Nx=B can't have a solution.

7. Apr 12, 2010

### jumbogala

That makes a lot of sense. I understand now, thank you!