Linear Algebra Change Matrices Confusion

[jumbogala]

Homework Statement

Suppose N is an invertible n x n matrix, and let D = {f₁, f₂, ... , f_n} where f_i is column i of N for each i. If B is the standard basis of Rⁿ, show that M_BD(1_Rⁿ) = N.

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

Homework Equations

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 ... f_1n
: :
: :
f_n1 ... f_nn]

1Rⁿ(f₁) = f₁, 1_Rⁿ(f_n) = f_n.

f₁ can be written as f₁₁E₁ + ... + f_n1E_n
f_n can be written as f_n1E₁ + ... + f_nnE_n

Then M_BD(1_Rⁿ) 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?

 

[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.

 

[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.

 

[Dick]

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?

 

[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...

 

[Dick]

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...

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.

 

[jumbogala]

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