How Do You Change the Basis of an Operator in Linear Algebra?

[perplexabot]

Homework Statement

Homework Equations

[itex] \check{T} = BTB^{-1} [/itex] (eq1)

The Attempt at a Solution

Ok, so I have a couple of questions here if I may ask... First, I want to be sure I understand the wording of (a) and (b) correctly. Is the following true?:

(a)

Write the matrix T in the basis of x1,x2,x3.
Write the matrix T in the standard R4 basis.
(I just want to make sure that a [and b] actually has two separate parts to be solved!)​

Finally, my attempt:

So, I am assuming that the T is already in the x1,x2,x3 basis, is that right?
Now to perform a change of basis on T to the standard [itex] R^{4} [/itex] basis.
This is where I am running into a problem!

I know that to get B of eq1 I must perform the operator on the new set of basis (the standard [itex] R^{4} [/itex] basis in this case), use the results to write the new basis in terms of the old one, then the column space of this solution will be B. Once again, assuming to this point what I have done is correct, I do the following:

Let {ε1,ε2, ε3, ε4} be the four standard basis of [itex] R^{4} [/itex]
T*ε1 =
T being a 4 x 3 matrix and ε1 a 4 x 1 matrix, I am not able to do the multiplication and everything fails!

I have really been at this question for a long long time. Any help/hints will be greatly appreciated. Thank you for your time.

 

[RUber]

I think what you have already done is to write T in the standard basis for ##\mathbb{R}^3## and ##\mathbb{R}^4##.
To do this in the ##\{x_1, x_2, x_3\}## basis, you would be looking for something that takes
##1x_1= \left[ \begin{array}{c} 1\\ 0\\ 0 \end{array}\right] ## to ##y_1##. And so on for ##x_2, x_3##.

 

[perplexabot]

I think what you have already done is to write T in the standard basis for ##\mathbb{R}^3## and ##\mathbb{R}^4##.

So you are saying the T that I have found is in the basis for ##\mathbb{R}^3## AND ##\mathbb{R}^4##?

To do this in the ##\{x_1, x_2, x_3\}## basis, you would be looking for something that takes
##1x_1= \left[ \begin{array}{c} 1\\ 0\\ 0 \end{array}\right] ## to ##y_1##. And so on for ##x_2, x_3##.

Wow. I think I understand what to do. So just form T from scratch instead of going through the change of basis. That makes so much sense. I will try that now. THANKS!

 

[RUber]

Right, both outputs are in the standard basis for ##\mathbb{R}^4##.
If you are using ##\{x_i\}## as your basis vectors, then ##\{x_1,x_2,x_3\} ## should look like the 3x3 identity matrix.
And T in this basis would be the operator that takes the 3x3 identity to ##\{y_1,y_2,y_3\} ##

 

[perplexabot]

Right, both outputs are in the standard basis for ##\mathbb{R}^4##.
If you are using ##\{x_i\}## as your basis vectors, then ##\{x_1,x_2,x_3\} ## should look like the 3x3 identity matrix.
And T in this basis would be the operator that takes the 3x3 identity to ##\{y_1,y_2,y_3\} ##

Thank you, that makes great sense.
EDIT: So T = y