Proving Similar Matrices with Examples and Step-by-Step Instructions

[holezch]

Homework Statement

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.

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

 

[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 [L_A]_gamma means?

 

[holezch]

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 [L_A]_gamma means?

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 [L_A]_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!

 

[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.

 

[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 :)

 

[hgfalling]

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