# Similar matrices help

1. Jul 23, 2010

### holezch

1. The problem statement, all variables and given/known data
Let A $$\in M_{n x n }(F)$$ and let $$\gamma$$ be an ordered basis for $$F^{n}$$. Then $$[L_{A}]_{\gamma} = Q^{-1} A Q$$, where Q is the n x n matrix whose j-th column is the j-th vector of gamma.

3. The attempt at a solution

I think I'm confused about some of the technical details.. :S And I don't really have much of a plan for proving this at all :S.. any help? thank you

2. Jul 23, 2010

### hgfalling

Well, you'll have to show us something.
Do you understand all the notation of the statement?

Maybe writing it out in words will help. Here's one way of writing the first sentence:

Let A be an element of the set of nxn matrices with elements in the field F.

Of particular interest: Do you understand what [LA]gamma means?

3. Jul 23, 2010

### holezch

I think the notation may be messing with me.. I'm also confused about the changing bases.. We need to change basis x to basis gamma. Are we just not given the basis x?

and [LA]gamma is a linear transformation, so we need to decide the coordinate vectors A* x_i where x _ i are the vectors from gamma
thanks!

4. Jul 23, 2010

### hgfalling

The original basis here is otherwise unspecified. Just assume that it's some other basis (besides $\gamma$).

As you say, $\left[ L_A \right]_\gamma$ is the linear transformation corresponding to the matrix A in the original basis, but with respect to the basis $\gamma$ instead.

If you were in the original basis and somebody came up with a vector u and asked you to do the linear transformation, you could do that easily, right? Just give them back Au. Now someone is coming up to you with a vector in the new basis v, and wants the result of applying the linear transformation in the new basis ($\gamma$). So if we can take v back to the old basis, apply A, then bring the result back to the new basis, we will have their answer.

This is where Q comes in. So now, show that Q-1 and Q do these things.

5. Jul 23, 2010

### holezch

thanks, I have another question: it says that the j-th column of Q is the j-th vector of gamma.. but Q is a matrix , so it's the jth vector of gamma under what basis coordinates?
the matrix Q would be in the coordinates of B' if Q is B -> B', so are we looking for gamma vectors with respect to gamma ? thanks :)

6. Jul 23, 2010

### hgfalling

The vectors in $\gamma$ would be in terms of the old basis.